Chapter 24 - Application of Integrals

In the previous section, we completed a discussion on integrals. In this chapter, we will see application of integrals.

Area under simple curves

One important application of integrals is the calculation of area under simple curves. Some basics about this application, can be written in steps:

1. In our earlier classes, we have seen the methods for finding areas of simple geometric figures like squares, rectangles, triangles, trapeziums etc.,  
• In those simple geometric figures, all sides are straight lines.

2. Now consider fig.24.1 below:

Fig.24.1

• AB, BC and CD are straight lines.
• But AD is a curve. AD is portion of the graph of the function $\small{f(x) = x^3 + 2x + 3}$
• In such a situation, we use integral calculus to find the area ABCDA.

3. Consider fig.24.2(a) below:

Fig.24.2

• Recall how we calculated the area bounded by the four items:
    ♦ The curve y = f(x)
    ♦ The vertical line x = a
    ♦ The vertical line x = b
    ♦ The horizontal line y=0 (x-axis)
• We assumed that, the required area is composed of infinite number of very thin strips of width dx. The sum of the areas of all those infinite strips will give the required area. One such sample strip is shown in the fig.24.2(a).
• This sample strip is situated at a distance of ‘x’ from the origin. So height of this sample strip will be y=f(x). Consequently, area of this sample strip will be ydx, which is same as f(x) dx.
• Note that, since there are infinite strips, the width dx is infinitesimal.
• Area of the sample strip will be so small that, we denote the area as dA. So dA = f(x) dx.
• Symbolically, we denoted the sum of all the strips as: $\small{\int_a^b{\left[f(x) \right]dx}}$
• So we can write: $\small{A~=~\int_a^b{\left[f(x) \right]dx}~=~\int_a^b{\left[dA \right]}}$

4. Consider fig.24.2(b) above.
• Here also, the area is bounded by the same four items:
    ♦ The curve y = f(x)
    ♦ The vertical line x = a
    ♦ The vertical line x = b
    ♦ The horizontal line y=0 (x-axis)
• We are tempted to apply the same equation: $\small{A~=~\int_a^b{\left[f(x) \right]dx}}$
But the calculated area will be −ve. We know that, area has no sign. So in such a situation, we must take the absolute value. We can write:
$\small{A~=~\left|\int_a^b{\left[f(x) \right]dx} \right|}$

5. Consider fig.24.3(a) below:

Fig.24.3

• The area is bounded by the four items:
    ♦ The curve x = g(x)
    ♦ The horizontal line y = c
    ♦ The horizontal line y = d
    ♦ The vertical line x=0 (y-axis)

• Here we can modify the equation that we saw in step (3) above. We get: $\small{A~=~\int_c^d{\left[g(y) \right]dy}~=~\int_c^d{\left[dA \right]}}$ 

6. Consider fig.24.3(b) above.
• Here we are tempted to apply the same equation as in step (5) above: $\small{A~=~\int_c^d{\left[g(y) \right]dy}}$
But the calculated area will be −ve. We know that, area has no sign. So in such a situation, we must take the absolute value. We can write:
$\small{A~=~\left|\int_c^d{\left[g(y) \right]dy} \right|}$

7. Consider fig.24.4 below:

Fig.24.4

• We want the area is bounded by the four items:
    ♦ The curve y = f(x)
    ♦ The vertical line x = a
    ♦ The vertical line x = b
    ♦ The horizontal line y=0 (x-axis)

• Such an area is the sum of two areas:
   ♦ violet area (A₁)
   ♦ yellow area (A₂)
• If we use the equation $\small{A~=~\int_a^b{\left[f(x) \right]dx}}$, we will be getting $\small{-A_1~+~A_2}$
• Two avoid this error, we calculate A₁ and A₂ separately. Then we find the sum: A = |A₁| + A₂

Now we will see a solved example

Solved example 24.1
Find the area enclosed by the circle $\small{x^2~+~ y^2~=~a^2}$
Solution:
1. First we write the given equation in the form: y = f(x)
$\small{x^2~+~ y^2~=~a^2}$

$\small{\Rightarrow y^2~=~a^2~-~x^2}$

$\small{\Rightarrow y~=~f(x)~=~\pm \sqrt{a^2~-~x^2}}$

• In the above equation, if we input all x values from the interval [−a,a], we will get the ordered pairs, which when plotted, will give the red circle in fig.24.5 below:

Fig.24.5

2. Consider fig.24.5(a) above. We want the area of the portion shaded in violet color. This portion is named as OABOA.

• The portion shaded in violet color is bounded by four items:
    ♦ The curve y = f(x)
    ♦ The vertical line x = 0 (y-axis)
    ♦ The vertical line x = a
    ♦ The horizontal line y=0 (x-axis)

• Consider the thin vertical strip of width dx. This strip is situated at a distance of x from O. So height of the strip will be f(x). So area of the strip will be f(x) dx.

Therefore, area (A) of the violet portion can be obtained as:
$\small{A~=~\int_0^a{\left[f(x) \right]dx}~=~\int_0^a{\left[\sqrt{a^2~-~x^2} \right]dx}}$

• Here we discard $\small{f(x)~=~-\sqrt{a^2~-~x^2}}$  and take $\small{f(x)~=~+\sqrt{a^2~-~x^2}}$. This is because, OABOA is in the first quadrant. In this quadrant, all y values are +ve.

3. This integration can be done as shown below:

• First we find the indefinite integral F:

$\small{F~=~\int{\left[\sqrt{a^2~-~x^2} \right]dx}}$

• This is a standard integral (see section 23.19). We have:

$\small{F~=~\int{\left[\sqrt{a^2~-~x^2} \right]dx}~=~\frac{x}{2} \sqrt{a^2~-~x^2}~+~\frac{a^2}{2} \sin^{-1}\frac{x}{a}}$

Therefore, $\small{A~=~\left[\frac{x}{2} \sqrt{a^2~-~x^2}~+~\frac{a^2}{2} \sin^{-1}\frac{x}{a} \right]_0^{a}}$

$\small{~=~\left[\frac{a}{2} \sqrt{a^2~-~a^2}~+~\frac{a^2}{2} \sin^{-1}\frac{a}{a} \right]~-~\left[\frac{0}{2} \sqrt{a^2~-~0^2}~+~\frac{a^2}{2} \sin^{-1}\frac{0}{a} \right]}$

$\small{~=~\left[0~+~\frac{a^2}{2} \sin^{-1}1 \right]~-~\left[0~+~\frac{a^2}{2} \sin^{-1}0 \right]}$

$\small{~=~\left[\frac{a^2}{2} \left(\frac{\pi}{2} \right) \right]~-~\left[0~+~0 \right]~=~\frac{\pi a^2}{4}}$

4. In the fig.24.5(a) above, we see that, the circle is symmetrical about both x and y axes. So the total area of the circle is four times the area of OABOA.

• We can write:
Whole area enclosed by the circle
$\small{~=~(4)\frac{\pi \,a^2}{4}~=~\pi\,a^2}$

• Note that, this result is in full agreement with the well known formula: Area of a circle = $\small{\pi r^2}$.
In our present case, the radius is 'a'. 

Alternate method:

1. First we write the given equation in the form: x = f(y)

$\small{x^2~+~ y^2~=~a^2}$

$\small{\Rightarrow x^2~=~a^2~-~y^2}$

$\small{\Rightarrow x~=~f(y)~=~\pm \sqrt{a^2~-~y^2}}$

• In the above equation, if we input all y values from the interval [−a,a], we will get the ordered pairs, which when plotted, will give the red circle in fig.24.5(a) and (b) above.

2. Consider fig.24.5(b) above. We want the area of the portion shaded in blue color. This portion is named as OBCOB.

• The portion shaded in blue color is bounded by four items:
    ♦ The curve x = f(y)
    ♦ The horizontal line y = 0 (x-axis)
    ♦ The horizontal line y = a
    ♦ The vertical line x = 0 (y-axis)
    
• Consider the thin horizontal strip of thickness dy. This strip is situated at a distance of y from O. So length of the strip will be f(y). So area of the strip will be f(y) dy.

Therefore, area (A) of the blue portion can be obtained as:
$\small{A~=~\int_0^a{\left[f(y) \right]dy}~=~\int_0^a{\left[-\sqrt{a^2~-~y^2} \right]dy}}$

• Here we discard $\small{x~=~f(y)~=~\sqrt{a^2~-~y^2}}$  and take $\small{x~=~f(y)~=~-\sqrt{a^2~-~y^2}}$. This is because, OBCOB is in the second quadrant. In this quadrant, all x values are −ve.

3. This integration can be done as shown below:

• First we find the indefinite integral F:

$\small{F~=~\int{\left[-\sqrt{a^2~-~y^2} \right]dy}~=~(-1)\int{\left[\sqrt{a^2~-~y^2} \right]dy}}$

• This is a standard integral (see section 23.19). We have:

$\small{F~=~(-1)\int{\left[\sqrt{a^2~-~y^2} \right]dy}~=~(-1)\left[\frac{x}{2} \sqrt{a^2~-~y^2}~+~\frac{a^2}{2} \sin^{-1}\frac{y}{a} \right]}$
Therefore, $\small{A~=~(-1)\left[\frac{y}{2} \sqrt{a^2~-~y^2}~+~\frac{a^2}{2} \sin^{-1}\frac{y}{a} \right]_0^{a}}$

$\small{~=~(-1)\left[\frac{a}{2} \sqrt{a^2~-~a^2}~+~\frac{a^2}{2} \sin^{-1}\frac{a}{a} \right]~-~(-1)\left[\frac{0}{2} \sqrt{a^2~-~0^2}~+~\frac{a^2}{2} \sin^{-1}\frac{0}{a} \right]}$

$\small{~=~(-1)\left[0~+~\frac{a^2}{2} \sin^{-1}1 \right]~+~\left[0~+~\frac{a^2}{2} \sin^{-1}0 \right]}$

$\small{~=~(-1)\left[\frac{a^2}{2} \left(\frac{\pi}{2} \right) \right]~+~\left[0~+~0 \right]~=~-\frac{\pi a^2}{4}}$

• Area cannot be −ve. We must take the absolute value. So we get:

$\small{A~=~ \left|-\frac{\pi a^2}{4} \right|~=~\frac{\pi a^2}{4}}$

4. Since the circle is symmetrical about both x and y axes, total area of the circle is four times the area of OBCOB.

• We can write:
Whole area enclosed by the circle
$\small{~=~(4)\frac{\pi \,a^2}{4}~=~\pi\,a^2}$

• We get the same result in both methods. For this problem, there is a total of eight methods (vertical or horizontal strip in each of the four quadrants) available. The reader may try all the eight methods to get a good understanding about the process.

Solved example 24.2
Find the area enclosed by the ellipse $\small{\frac{x^2}{a^2}~+~ \frac{y^2}{b^2}~=~1}$
Solution:
1. First we write the given equation in the form: y = f(x)
$\small{\frac{x^2}{a^2}~+~ \frac{y^2}{b^2}~=~1}$

$\small{\Rightarrow \frac{y^2}{b^2}~=~1~-~\frac{x^2}{a^2}}$

$\small{\Rightarrow y^2~=~b^2 \left(1~-~\frac{x^2}{a^2} \right)~=~b^2 \left(\frac{a^2~-~x^2}{a^2} \right)~=~\frac{b^2}{a^2}\left(a^2~-~x^2 \right)}$

$\small{\Rightarrow y~=~f(x)~=~\pm \frac{b}{a} \sqrt{a^2~-~x^2}}$

• In the above equation, if we input all x values from the interval [−a,a], we will get the ordered pairs, which when plotted, will give the red ellipse in fig.24.6 below:

Fig.24.6

2. Consider fig.24.6(a) above. We want the area of the portion shaded in violet color. This portion is named as OABOA.

• The portion shaded in violet color is bounded by four items:
    ♦ The curve y = f(x)
    ♦ The vertical line x = 0 (y-axis)
    ♦ The vertical line x = a
    ♦ The horizontal line y=0 (x-axis)

• Consider the thin vertical strip of width dx. This strip is situated at a distance of x from O. So height of the strip will be f(x). So area of the strip will be f(x) dx.

Therefore, area (A) of the violet portion can be obtained as:
$\small{A~=~\int_0^a{\left[f(x) \right]dx}~=~\int_0^a{\left[\frac{b}{a} \sqrt{a^2~-~x^2} \right]dx}}$

• Here we discard $\small{f(x)~=~-\frac{b}{a} \sqrt{a^2~-~x^2}}$  and take $\small{f(x)~=~+\frac{b}{a} \sqrt{a^2~-~x^2}}$. This is because, OABOA is in the first quadrant. In this quadrant, all y values are +ve.

3. This integration can be done as shown below:

• First we find the indefinite integral F:

$\small{F~=~\frac{b}{a} \int{\left[\sqrt{a^2~-~x^2} \right]dx}}$

• This is a standard integral (see section 23.19). We have:

$\small{F~=~\int{\left[\sqrt{a^2~-~x^2} \right]dx}~=~\frac{x}{2} \sqrt{a^2~-~x^2}~+~\frac{a^2}{2} \sin^{-1}\frac{x}{a}}$
Therefore, $\small{A~=~\left[\frac{x}{2} \sqrt{a^2~-~x^2}~+~\frac{a^2}{2} \sin^{-1}\frac{x}{a} \right]_0^{a}}$

$\small{~=~\frac{b}{a} \left[\frac{a}{2} \sqrt{a^2~-~a^2}~+~\frac{a^2}{2} \sin^{-1}\frac{a}{a} \right]~-~\frac{b}{a} \left[\frac{0}{2} \sqrt{a^2~-~0^2}~+~\frac{a^2}{2} \sin^{-1}\frac{0}{a} \right]}$

$\small{~=~\frac{b}{a} \left[0~+~\frac{a^2}{2} \sin^{-1}1 \right]~-~\frac{b}{a} \left[0~+~\frac{a^2}{2} \sin^{-1}0 \right]}$

$\small{~=~\frac{b}{a} \left[\frac{a^2}{2} \left(\frac{\pi}{2} \right) \right]~-~\frac{b}{a} \left[0~+~0 \right]~=~\frac{\pi ab}{4}}$

4. In the fig.24.6(a) above, we see that, the ellipse is symmetrical about both x and y axes. So the total area of the ellipse is four times the area of OABOA.

• We can write:
Whole area enclosed by the ellipse
$\small{~=~(4)\frac{\pi \,ab}{4}~=~\pi\,ab}$

Alternate method:

1. First we write the given equation in the form: x = f(y)
$\small{\frac{x^2}{a^2}~+~ \frac{y^2}{b^2}~=~1}$

$\small{\Rightarrow \frac{x^2}{a^2}~=~1~-~\frac{y^2}{b^2}}$

$\small{\Rightarrow x^2~=~a^2 \left(1~-~\frac{y^2}{b^2} \right)~=~a^2 \left(\frac{b^2~-~y^2}{b^2} \right)~=~\frac{a^2}{b^2}\left(b^2~-~y^2 \right)}$

$\small{\Rightarrow x~=~f(y)~=~\pm \frac{a}{b} \sqrt{b^2~-~y^2}}$

• In the above equation, if we input all y values from the interval [−b,b], we will get the ordered pairs, which when plotted, will give the red ellipse in fig.24.6 above.

2. Consider fig.24.6(a) above. We want the area of the portion shaded in blue color. This portion is named as OA'B'OA'.

• The portion shaded in blue color is bounded by four items:
    ♦ The curve y = f(x)
    ♦ The horizontal line y = 0 (x-axis)
    ♦ The horizontal line y = −b
    ♦ The vertical line x = 0 (y-axis)

• Consider the thin horizontal strip of thickness dy. This strip is situated at a distance of y from O. So length of the strip will be f(y). So area of the strip will be f(y) dy.

Therefore, area (A) of the blue portion can be obtained as:
$\small{A~=~\int_{-b}^0{\left[f(y) \right]dy}~=~\int_{-b}^0{\left[-\frac{a}{b} \sqrt{b^2~-~y^2} \right]dy}}$

• Here we discard $\small{f(y)~=~+\frac{a}{b} \sqrt{b^2~-~y^2}}$  and take $\small{f(y)~=~-\frac{a}{b} \sqrt{b^2~-~y^2}}$. This is because, OA'B'OA' is in the third quadrant. In this quadrant, all x values are −ve.

3. This integration can be done as shown below:

• First we find the indefinite integral F:

$\small{F~=~-\frac{a}{b} \int{\left[\sqrt{b^2~-~y^2} \right]dy}}$

• This is a standard integral (see section 23.19). We have:

$\small{F~=~\int{\left[\sqrt{b^2~-~y^2} \right]dy}~=~\frac{y}{2} \sqrt{b^2~-~y^2}~+~\frac{b^2}{2} \sin^{-1}\frac{y}{b}}$
Therefore, $\small{A~=~\left[\frac{y}{2} \sqrt{b^2~-~y^2}~+~\frac{b^2}{2} \sin^{-1}\frac{y}{b} \right]_{-b}^{0}}$

$\small{~=~\left(-\frac{a}{b} \right) \left[\frac{0}{2} \sqrt{b^2~-~0^2}~+~\frac{b^2}{2} \sin^{-1}\frac{0}{b} \right]~-~\left(-\frac{a}{b} \right) \left[\frac{-b}{2} \sqrt{b^2~-~(-b)^2}~+~\frac{b^2}{2} \sin^{-1}\frac{-b}{b} \right]}$

$\small{~=~\left(-\frac{a}{b} \right) \left[0~+0 \right]~+~\left(\frac{a}{b} \right) \left[0~+~\frac{b^2}{2} \sin^{-1}(-1) \right]~~\because \sin^{-1}(-1)~=~-\sin^{-1}(1) }$

$\small{~=~0~+~\left(\frac{a}{b} \right) \left[-\frac{b^2}{2} \sin^{-1}(1) \right]}$

$\small{~=~\frac{a}{b} \left[-\frac{b^2}{2} \left(\frac{\pi}{2} \right) \right]~=~\frac{-\pi ab}{4}}$

• Area cannot be −ve. We must take the absolute value. So we get:

$\small{A~=~ \left|-\frac{\pi ab}{4} \right|~=~\frac{\pi ab}{4}}$

4. In the fig.24.6(b) above, we see that, the ellipse is symmetrical about both x and y axes. So the total area of the ellipse is four times the area of OA'B'OA'.

• We can write:
Whole area enclosed by the ellipse
$\small{~=~(4)\frac{\pi \,ab}{4}~=~\pi\,ab}$

• We get the same result in both methods. For this problem, there is a total of eight methods (vertical or horizontal strip in each of the four quadrants) available. The reader may try all the eight methods to get a good understanding about the process.

---

In the next section, we will see area bounded by a curve and a line.

Previous

Contents

Next 

Copyright©2025 Higher secondary mathematics.blogspot.com

23.31 - Miscellaneous Examples (3) on Integrals

In the previous section, we saw some miscellaneous examples on integrals. In this section, we will see a few more miscellaneous examples.

Solved Example 23.148
Integrate $\small{\frac{e^x}{(1+e^x)(2+e^x)}}$    
Solution:
1. Let us put $\small{u = 1+e^x}$

• Then $\small{\frac{du}{dx}~=~e^x \Rightarrow e^x\,dx~=~du}$

• Also, since $\small{u = 1+e^x}$, we can write:

$\small{2+e^x = u+1}$

2. So we want:

$\small{I = \int{\left[\frac{e^x}{(1+e^x)(2+e^x)}\right]dx}= \int{\left[\frac{1}{(u)(u+1)}\right]du}= \int{\left[\frac{1}{u^2 + u}\right]du}}$

3. This is a standard integral of the form:

$\small{ \int{\left[\frac{px + q}{ax^2 + bx + c}\right]du}}$, where p = 0, q = 1, a = 1, b = 1 and c = 0

We get: $\small{I~=~\int{\left[\frac{1}{u^2 + u}\right]du}~=~(-1)\log \left| \frac{1}{u}~+~1 \right|~+~\rm{C}}$

$\small{~=~(-1)\log \left| \frac{1+u}{u} \right|~+~\rm{C}~=~\log \left| \frac{u}{1+u} \right|~+~\rm{C}}$

4. Substituting for u, we get:

$\small{I~=~\log \left| \frac{1+e^x}{2+e^x} \right|~+~\rm{C}}$

Solved Example 23.149
Integrate $\small{f'(ax+b) \left[f(ax+b) \right]^n}$
Solution:
1. Let us put $\small{u = f(ax+b)}$

• Then $\small{\frac{du}{dx}~=~a\,f'(ax+b) \Rightarrow a\,f'(ax+b)\,dx~=~du}$

2. So we want:

$\small{I = \int{\left[f'(ax+b) \left[f(ax+b) \right]^n\right]dx}= \int{\left[\frac{(a)f'(ax+b) \left[f(ax+b) \right]^n}{a}\right]dx}}$

$\small{= \int{\left[\frac{u^n}{a}\right]du}}$

3. This integration gives: $\small{I = \frac{u^{n+1}}{a(n+1)}~+~\rm{C}}$

4. Substituting for u, we get:

$\small{I = \frac{[f(ax+b]^{n+1}}{a(n+1)}~+~\rm{C}}$

Solved Example 23.150
Integrate $\small{\left(\frac{2 + \sin(2x)}{1 + \cos(2x)} \right)e^x}$
Solution:
1. 1. First we will rearrange the given expression:

$\small{\left(\frac{2 + \sin(2x)}{1 + \cos(2x)} \right)e^x~=~\left(\frac{2 + 2\sin(x) \cos (x)}{2 \cos^2 x} \right)e^x}$

$\small{~=~\left(\frac{1}{\cos^2 x}~+~\tan x \right)e^x~=~\left(\sec^2 x~+~\tan x \right)e^x}$

2. So we want:

$\small{I = \int{\left[\left(\sec^2 x~+~\tan x \right)e^x\right]dx}}$

• $\small{\sec^2 x}$ is the derivative of $\small{\tan x}$. So this is of the form:

$\small{\int{\left[\left(f(x)~+~f'(x) \right)e^x\right]dx}~=~e^x\,f(x)~+~\rm{C}}$

3. Thus we get:

$\small{I = \int{\left[\left(\frac{2 + \sin(2x)}{1 + \cos(2x)} \right)e^x\right]dx}= \int{\left[\left(\sec^2 x~+~\tan x \right)e^x\right]dx}~=~e^x\,\tan(x)~+~\rm{C}}$

Solved Example 23.151
Integrate $\small{\frac{\sqrt{x^2 + 1}\left[\log(x^2 + 1)~-~2 \log x \right]}{x^4}}$
Solution:
1. First we will rearrange the given expression:

Put $\small{x~=~\tan u}$

$\small{\Rightarrow 1+x^2~=~1+\tan^2 u~=~\sec^2 u}$

• Also, $\small{\frac{dx}{du}~=~\sec^2 u \Rightarrow dx~=~\sec^2 u \, du}$

• So we can write:

$\small{I~=~\int{\left[\frac{\sqrt{x^2 + 1}\left[\log(x^2 + 1)~-~2 \log x \right]}{x^4} \right]dx}~=~\int{\left[\frac{\sqrt{\sec^2 u}\left[\log(\sec^2 u)~-~ \log (\tan^2 u) \right]}{\tan^4 u} \right]\sec^2 u\,du}}$

$\small{~=~\int{\left[\frac{\left[\log\left(\frac{\sec^2 u}{\tan^2 u} \right) \right]}{\tan^4 u} \right]\sec^3 u\,du}~=~\int{\left[\frac{\left[\log\left(\frac{1}{\sin^2 u} \right) \right]}{\tan^4 u\left(\cos^3 u \right)} \right]du}}$

$\small{~=~\int{\left[\frac{\cos u\left[\log\left(\frac{1}{\sin^2 u} \right) \right]}{\sin^4 u} \right]du}}$

2. We will use the method of integration by parts.

(a) Assigning first and second functions:

   ♦ Let first function be: $\small{f(u)=\log\left(\frac{1}{\sin^2 u} \right)}$

   ♦ Let second function be: $\small{g(u)=\frac{\cos u}{\sin^4 u}}$

(b) Finding A:

$\small{A~=~\int{\left[g(u) \right]dx}~=~\frac{- 1}{3\sin^3 u}}$

(c) $\small{\big[f(u) \left(A \right) \big]~=~\big[ \log\left(\frac{1}{\sin^2 u} \right)\big] \,\big[\frac{- 1}{3 \sin^3 u} \big]}$

• This is the first term.

(d) $\small{f'(u)~=~\frac{-2 \cos(u)}{\sin(u)}}$

(e) $\small{\int{\big[f'(u)\,\left(A \right)  \big]du}~=~\int{\big[\frac{-2 \cos(u)}{\sin(u)}\,\left(\frac{- 1}{3 \sin^3 (u)} \right)\big]du}}$

$\small{~=~\frac{2}{3}\int{\big[\frac{\cos(u)}{\sin^4(u)}\big]du}~=~\frac{-2}{9\sin^3 u}}$

• This is the second term.

(f) So we get:

$\small{I~=~\text{First term - Second term}}$

$\small{~=~\big[ \log\left(\frac{1}{\sin^2 u} \right)\big] \,\big[\frac{- 1}{3 \sin^3 u} \big]~-~\big[\frac{-2}{9 \sin^3 u} \big]}$

$\small{~=~\big[ \log\left(\frac{1}{\sin^2 u} \right)\big] \,\big[\frac{- 1}{3 \sin^3 u} \big]~+~\big[\frac{2}{9 \sin^3 u} \big]}$

$\small{~=~\big[\frac{2}{9 \sin^3 u} \big]~-~\big[ \log\left(\frac{1}{\sin^2 u} \right)\big] \,\big[\frac{1}{3 \sin^3 u} \big]}$

$\small{~=~\big[\frac{1}{3\sin^3 u} \big]\bigg[\frac{2}{3}~-~ \log\left(\frac{1}{\sin^2 u} \right)\bigg]}$

3. Now we will substitute for u:

(a) We wrote $\small{x~=~\tan u}$

$\small{\Rightarrow 1+x^2~=~1+\tan^2 u~=~\sec^2 u}$

$\small{\Rightarrow \cos^2 u~=~\frac{1}{1+x^2}}$

$\small{\Rightarrow \sin^2 u~=~1~-~\frac{1}{1+x^2}~=~\frac{x^2}{1+x^2}}$

$\small{\Rightarrow \frac{1}{\sin^2 u}~=~1~+~\frac{1}{x^2}}$

$\small{\Rightarrow \frac{1}{\sin u}~=~\left(1~+~\frac{1}{x^2} \right)^{1/2}}$

$\small{\Rightarrow \frac{1}{\sin^3 u}~=~\left(1~+~\frac{1}{x^2} \right)^{3/2}}$

(b) So we get:

$\small{I~=~\big[\frac{1}{3\sin^3 u} \big]\bigg[\frac{2}{3}~-~ \log\left(\frac{1}{\sin^2 u} \right)\bigg]}$

$\small{~=~\big[\frac{1}{3} \left(1~+~\frac{1}{x^2} \right)^{3/2} \big]\bigg[\frac{2}{3}~-~ \log\left(1~+~\frac{1}{x^2} \right)\bigg]~+~\rm{C}}$

Solved Example 23.152
Integrate $\small{\frac{1}{e^x~+~e^{-x}}}$
Solution:
1. First we will rearrange the given expression:

$\small{\frac{1}{e^x~+~e^{-x}}~=~\frac{1}{e^x~+~\frac{1}{e^x}}~=~\frac{e^x}{e^{2x}~+~1}}$

Put $\small{u~=~e^x}$

• Then, $\small{\frac{du}{dx}~=~e^x \Rightarrow e^x\,dx ~=~du}$

2. So we can write:

$\small{I~=~\int{\left[\frac{1}{e^x~+~e^{-x}} \right]dx}~=~\int{\left[\frac{e^x}{e^{2x}~+~1} \right]du}~=~\int{\left[\frac{1}{u^{2}~+~1} \right]du}}$

3. This is a standard integral. We get:

$\small{I~=~\tan^{-1}u~+~\rm{C}}$

4. Substituting for u, we get:

$\small{I~=~\tan^{-1}(e^x)~+~\rm{C}}$

Solved Example 23.153
Evaluate $\small{\int_0^1{\left[e^{2 - 3x} \right]dx}}$ as a limit of a sum.
Solution:
1. 1. $\small{\int_{0}^{1}{\left[e^{2-3x} \right]dx}}$ is the area bounded by the four items:
   ♦ The curve $\small{y = f(x) = e^{2-3x}}$
   ♦ The vertical line x = 0 (y-axis)
   ♦ The vertical line x = 1
   ♦ The horizontal line y = 0 (x-axis)
   
2. We have:
$\small{\int_a^b{\left[f(x) \right]dx}~=~\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~.~.~.~+~f(a+nh) \right]}$

• In our present case, a = 0 and b = 1

So $\small{h~=~\frac{b-a}{n}~=~\frac{1-0}{n}~=~\frac{1}{n}}$

3. Now the formula becomes:
Area =
$\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~f(a+3h)~+~~.~.~.~+f(a+nh) \right]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{1}{n} \left[f(a+\frac{1}{n})~+~f(a+\frac{2}{n})~+~f(a+\frac{3}{n})~+~~.~.~.~+f(a+(n)\frac{1}{n}) \right]}$

$\small{~=~\lim_{n\rightarrow \infty} \frac{1}{n} \bigg[\big[e^{2-3(a+\frac{1}{n})}\big]~+~\big[e^{2-3(a+\frac{2}{n})}\big]~+~\big[e^{2-3(a+\frac{3}{n})}\big]~+~~.~.~.~}$

$\small{~~~~~~~~~.~.~.~+~\big[e^{2-3(a+(n)\frac{1}{n})}\big]\bigg]}$

4. Let us determine the quantity inside the large square brackets. We have to do a summation:

$\small{e^{2-3(a+\frac{1}{n})}~+~e^{2-3(a+\frac{2}{n})}~+~e^{2-3(a+\frac{3}{n})}~+~.~.~.~ \text{n terms}}$

$\small{~=~\left(e^2~\times~e^{-3a}~\times~e^{\frac{-3}{n}} \right)~+~\left(e^2~\times~e^{-3a}~\times~e^{\frac{-6}{n}} \right)~+~\left(e^2~\times~e^{-3a}~\times~e^{\frac{-9}{n}} \right)~+~.~.~.~ \text{n terms}}$

$\small{~=~e^{2-3a}\left(e^{\frac{-3}{n}}~+~e^{\frac{-6}{n}}~+~e^{\frac{-9}{n}}~+~.~.~.~ \text{n terms} \right)}$

$\small{~=~e^{2}\left(e^{\frac{-3}{n}}~+~e^{\frac{-6}{n}}~+~e^{\frac{-9}{n}}~+~.~.~.~ \text{n terms} \right)}$

$\small{~~~~~~\because {3a}~=~{3(0)}~=~0}$

• So inside the brackets, we have a geometric progression.

• First term = $\small{e^{\frac{-3}{n}}}$

• Common ratio  $\small{~r~=~\frac{e^{\frac{-6}{n}}}{e^{\frac{-3}{n}}}~=~\frac{e^{\frac{-9}{n}}}{e^{\frac{-6}{n}}}~=~e^{\frac{-3}{n}}}$

• Sum of the first n terms is given by: $\small{\frac{\text{First term}~\times~\left(r^{n-1}~-~1 \right)}{r~-~1}}$

$\small{r^{n-1}~=~\left(e^{\frac{-3}{n}}\right)^{n-1}~=~e^{\frac{-3n+3}{n}}~=~e^{-3+\frac{3}{n}}}$

• So we get:

Sum of all terms of the G.P

$\small{~=~\frac{e^{\frac{-3}{n}}~\times~\left({e^{-3+\frac{3}{n}}}~-~1 \right)}{e^{\frac{-3}{n}}~-~1}~=~\frac{{e^{-3}}~-~e^{\frac{-3}{n}} }{e^{\frac{-3}{n}}~-~1}}$

• So the summation is:

$\small{e^2 \left[\frac{{e^{-3}}~-~e^{\frac{-3}{n}} }{e^{\frac{-3}{n}}~-~1} \right]}$

5. So the limit in (3) becomes:

$\small{\lim_{n\rightarrow \infty} \frac{1}{n} \Bigg[e^2 \left[\frac{{e^{-3}}~-~e^{\frac{-3}{n}} }{e^{\frac{-3}{n}}~-~1} \right] \Bigg]}$

$\small{~=~e^2 \lim_{n\rightarrow \infty} \frac{1}{n}  \left[\frac{{e^{-3}}~-~e^{\frac{-3}{n}} }{e^{\frac{-3}{n}}~-~1} \right]}$

$\small{~=~e^2 \lim_{n\rightarrow \infty} \frac{1}{n}  \left[\frac{{e^{-3}} }{e^{\frac{-3}{n}}~-~1} \right]~-~e^2 \lim_{n\rightarrow \infty} \frac{1}{n}  \left[\frac{e^{\frac{-3}{n}} }{e^{\frac{-3}{n}}~-~1} \right]}$

$\small{~=~e^2 \lim_{n\rightarrow \infty} \frac{(-3)}{(-3)n}  \left[\frac{{e^{-3}} }{e^{\frac{-3}{n}}~-~1} \right]~-~e^2 \lim_{n\rightarrow \infty} \frac{(-3)}{(-3)n}  \left[\frac{e^{\frac{-3}{n}} }{e^{\frac{-3}{n}}~-~1} \right]}$

$\small{~=~\frac{e^2}{(-3)} \lim_{n\rightarrow \infty} \frac{-3}{n}  \left[\frac{{e^{-3}} }{e^{\frac{-3}{n}}~-~1} \right]~-~\frac{e^2}{(-3)} \lim_{n\rightarrow \infty} \frac{-3}{n}  \left[\frac{e^{\frac{-3}{n}} }{e^{\frac{-3}{n}}~-~1} \right]}$

$\small{~=~\frac{e^{-1}}{(-3)} \lim_{n\rightarrow \infty} \frac{-3}{n}  \left[\frac{{1} }{e^{\frac{-3}{n}}~-~1} \right]~-~\frac{e^2}{(-3)} \lim_{n\rightarrow \infty} \frac{-3}{n}  \left[\frac{e^{\frac{-3}{n}} }{e^{\frac{-3}{n}}~-~1} \right]}$

6. In the above expression, the limits can be evaluated as follows:

$\small{\frac{e^{-1}}{(-3)} \lim_{n\rightarrow \infty}   \left[\frac{{1} }{\frac{e^{\frac{-3}{n}}~-~1}{\frac{-3}{n}}} \right]
~-~\frac{e^2}{(-3)} \lim_{n\rightarrow \infty}   \left[\frac{{e^{\frac{-3}{n}}} }{\frac{e^{\frac{-3}{n}}~-~1}{\frac{-3}{n}}}  \right]}$

$\small{~=~\frac{e^{-1}}{(-3)}  \bigg[\frac{1}{1} \bigg]~-~\frac{e^2}{(-3)}\big[\frac{1}{1} \big]~=~\frac{e^{-1}~-~e^2}{(-3)}}$

$\small{~=~\frac{e^{2}~-~e^{-1}}{3}}$

• Here we use two facts:

(i) $\small{\lim_{n\rightarrow \infty}  \left[e^{\frac{-3}{n}} \right]~=~e^{\frac{-3}{\infty}}~=~e^0~=~1}$

(ii) Let $\small{\frac{-3}{n}~=~h}$.
Then $\small{h \rightarrow 0 ~\text{as}~n \rightarrow \infty}$

So $\small{\lim_{n\rightarrow \infty}  \Big[\frac{e^{\frac{-3}{n}}~-~1}{\frac{-3}{n}}\Big]~=~\lim_{n\rightarrow \infty}  \Big[\frac{e^{h}~-~1}{h}\Big]~=~1}$   

Solved Example 23.154
Evaluate $\small{\int_{0}^{\frac{\pi}{4}}{\left[\frac{\sin x \cos x}{\cos^4 x + \sin^4 x} \right]dx}}$
Solution:
1. Let us rearrange the given expression:

$\small{\frac{\sin x \cos x}{\cos^4 x + \sin^4 x}~=~\frac{\sin x \cos x}{\cos^4 x + \sin^4 x~+~2\sin^2x\,\cos^2x~-~2\sin^2x\,\cos^2x}}$

$\small{~=~\frac{\sin x \cos x}{\left(\cos^2 x + \sin^2 x \right)^2~-~2\sin^2x\,\cos^2x}~=~\frac{\sin x \cos x}{1~-~2\sin^2x\,\cos^2x}}$

$\small{~=~\frac{\sin x \cos x}{1~-~(4\sin^2x\,\cos^2x)/2}~=~\frac{2\sin x \cos x}{2~-~4\sin^2x\,\cos^2x}}$

$\small{~=~\frac{\sin (2x)}{2~-~\sin^2(2x)}~=~\frac{\sin (2x)}{1+[1~-~\sin^2(2x)]}~=~\frac{\sin (2x)}{1+[\cos^2(2x)]}}$

2. So we want: $\small{I~=~\int_{0}^{\frac{\pi}{4}}{\left[\frac{\sin x \cos x}{\cos^4 x + \sin^4 x} \right]dx}~=~\int_{0}^{\frac{\pi}{4}}{\left[\frac{\sin (2x)}{1+\cos^2(2x)} \right]dx}}$

3. First we will find the indefinite integral:

$\small{F~=~\int{\left[\frac{\sin (2x)}{1+\cos^2(2x)} \right]dx}}$

• Put $\small{u = \cos(2x)}$

Then $\small{\frac{du}{dx}~=~-2\sin(2x) \Rightarrow -2\sin(2x)\,dx~=~du}$

• We want:

$\small{F~=~\int{\left[\frac{(-2)\sin (2x)}{(-2)\left[1+\cos^2(2x) \right]} \right]dx}~=~\int{\left[\frac{1}{(-2)\left[1+u^2 \right]} \right]dx}~=~\frac{1}{(-2)}\int{\left[\frac{1}{1+u^2 } \right]dx}}$

• This is a standard integral. We get:

$\small{F~=~\left(\frac{-1}{2} \right)\tan^{-1}u}$

• Subsituting for u, we get:

$\small{F~=~\left(\frac{-1}{2} \right)\tan^{-1}\left[\cos(2x) \right]}$

4. Now we can evaluate the definite integral. We get:

$\small{I~=~\int_{0}^{\frac{\pi}{4}}{\left[\frac{\sin (2x)}{1+\cos^2(2x)} \right]dx}}$

$\small{~=~\big[\left(\frac{-1}{2} \right)\tan^{-1}\left[\cos(2x) \right]\big]_{0}^{\frac{\pi}{4}}}$

$\small{~=~\left(\frac{-1}{2} \right) \big[\tan^{-1}\left[\cos(2x) \right]\big]_{0}^{\frac{\pi}{4}}}$

$\small{~=~\left(\frac{-1}{2} \right) \big[\tan^{-1}\left[\cos\left(\frac{\pi}{2} \right) \right]~-~\tan^{-1}\left[\cos\left(0 \right) \right]\big]}$

$\small{~=~\left(\frac{-1}{2} \right) \big[\tan^{-1}\left[0 \right]~-~\tan^{-1}\left[1 \right]\big]}$

$\small{~=~\left(\frac{-1}{2} \right) \big[0~-~\frac{\pi}{4}\big]~=~\frac{\pi}{8}}$

Solved Example 23.155
Evaluate $\small{\int_{\frac{\pi}{2}}^{\pi}{\left[e^x\left(\frac{1-\sin x}{1 - \cos x} \right) \right]dx}}$
Solution:
1. Let us rearrange the portion inside braces:

$\small{\frac{1-\sin x}{1 - \cos x}~=~\frac{1-\sin x}{2 \sin^2\left(\frac{x}{2} \right)}~=~\frac{1}{2 \sin^2\left(\frac{x}{2} \right)}~-~\frac{\sin x}{2 \sin^2\left(\frac{x}{2} \right)}}$

$\small{~=~\frac{1}{2 \sin^2\left(\frac{x}{2} \right)}~-~\frac{2 \sin\left(\frac{x}{2} \right) \cos\left(\frac{x}{2} \right)}{2 \sin^2\left(\frac{x}{2} \right)}}$

$\small{~=~\frac{\csc^2\left(\frac{x}{2} \right)}{2}~-~\cot\left(\frac{x}{2} \right)~=~(-1)\left[\cot\left(\frac{x}{2} \right)~-~\frac{\csc^2\left(\frac{x}{2} \right)}{2} \right]}$

$\small{~=~(-1)\left[\cot\left(\frac{x}{2} \right)~+~\frac{(-1)\csc^2\left(\frac{x}{2} \right)}{2} \right]}$

2. So the given expression can be written as:

$\small{(-1)\,e^x\left[\cot\left(\frac{x}{2} \right)~+~\frac{(-1)\csc^2\left(\frac{x}{2} \right)}{2} \right]}$

• In the above result, $\small{\left[\frac{(-1)\csc^2\left(\frac{x}{2} \right)}{2} \right]}$ is the derivative of $\small{\left[\cot\left(\frac{x}{2} \right) \right]}$

• So the given expression can be written as:

$\small{(-1)\,e^x\left[f(x)~+~f'(x) \right]}$

3. So the indefinite integral can be written as:

$\small{F~=~(-1)\,e^x\left[f(x)\right]~=~(-1)\,e^x\left[\cot\left(\frac{x}{2} \right)\right]}$

4. Now we can evaluate the definite integral. We get:

$\small{I~=~\int_{\frac{\pi}{2}}^{\pi}{\left[e^x\left(\frac{1-\sin x}{1 - \cos x} \right) \right]dx}}$

$\small{~=~\big[(-1)\,e^x\left[\cot\left(\frac{x}{2} \right)\right]\big]_{\frac{\pi}{2}}^{\pi}}$

$\small{~=~\big[(-1)\,e^{\pi}\left[\cot\left(\frac{\pi}{2} \right)\right]\big]~-~\big[(-1)\,e^{\frac{\pi}{2}}\left[\cot\left(\frac{\pi}{4} \right)\right]\big]}$

$\small{~=~\big[(-1)\,e^{\pi}\left[0\right]\big]~-~\big[(-1)\,e^{\frac{\pi}{2}}\left[1\right]\big]}$

$\small{~=~e^{\frac{\pi}{2}}}$

Solved Example 23.156
Evaluate $\small{\int_{0}^{\frac{\pi}{2}}{\left[\frac{\cos^2 x}{\cos^2 x ~+~ 4 \sin^2 x} \right]dx}}$
Solution:
1. Let us rearrange the given expression:

$\small{\frac{\cos^2 x}{\cos^2 x ~+~ 4 \sin^2 x}~=~\frac{\cos^2 x}{\cos^2 x ~+~ 4(1-\cos^2 x)}~=~\frac{\cos^2 x}{4~-~3 \cos^2 x}}$

$\small{~=~\left(\frac{-1}{3} \right)\left[\frac{(-3)\cos^2 x}{4~-~3 \cos^2 x} \right]~=~\left(\frac{-1}{3} \right)\left[\frac{4~-~3\cos^2 x~-~4}{4~-~3 \cos^2 x} \right]}$

$\small{~=~\left(\frac{-1}{3} \right)\left[1~-~\frac{4}{4~-~3 \cos^2 x} \right]~=~\left(\frac{-1}{3} \right)\left[1~-~\frac{\frac{4}{\cos^2 x}}{\frac{4}{\cos^2 x}~-~\frac{3\cos^2 x}{\cos^2 x}} \right]}$

$\small{~=~\left(\frac{-1}{3} \right)\left[1~-~\frac{4 \sec^2 x}{4 \sec^2 x~-~3} \right]~=~\left(\frac{-1}{3} \right)\left[1~-~\frac{4 \sec^2 x}{4 (1+\tan^2 x)~-~3} \right]}$

$\small{~=~\left(\frac{-1}{3} \right)\left[1~-~\frac{4 \sec^2 x}{1~+~4 \tan^2 x} \right]}$ 

2. Now we can find the indefinite integral:

$\small{F~=~\int{\left[\frac{\cos^2 x}{\cos^2 x ~+~ 4 \sin^2 x} \right]dx}~=~\int{\bigg[\left(\frac{-1}{3} \right)\left[1~-~\frac{4 \sec^2 x}{1~+~4 \tan^2 x} \right] \bigg]dx}}$


$\small{~=~\int{\bigg[\left(\frac{-1}{3} \right) \bigg]dx}~+~\left(\frac{1}{3} \right)\int{\bigg[\frac{4 \sec^2 x}{1~+~4 \tan^2 x}  \bigg]dx}}$

$\small{~=~\frac{-x}{3}~+~\left(\frac{1}{3} \right)\int{\bigg[\frac{4 \sec^2 x}{1~+~4 \tan^2 x}  \bigg]dx}}$

$\small{~=~F_1~+~F_2}$

• $\small{F_2}$ can be calculated as follows:

• Put $\small{u~=~\tan x}$. Then $\small{\frac{du}{dx}~=~\sec^2 x \Rightarrow \sec^2 x\,dx~=~du}$

• So we get:

$\small{F_2~=~\left(\frac{1}{3} \right)\int{\bigg[\frac{4}{1~+~4(u)^2}  \bigg]dx}~=~\left(\frac{4}{3} \right)\int{\bigg[\frac{1}{1~+~(2u)^2}  \bigg]dx}}$

$\small{~=~\left(\frac{4}{3} \right)\left(\frac{1}{2} \right)\bigg[\tan^{-1}(2u)  \bigg]~=~\left(\frac{2}{3} \right)\bigg[\tan^{-1}(2 \tan x)  \bigg]}$

• Therefore, $\small{F~=~\frac{-x}{3}~+~\left(\frac{2}{3} \right)\left[\tan^{-1}(2 \tan x)  \right]}$

3. Now we can evaluate the definite integral. We get:

$\small{I~=~\big[\frac{-x}{3}~+~\left(\frac{2}{3} \right)\left[\tan^{-1}(2 \tan x)  \right]\big]_{0}^{\frac{\pi}{2}}}$

$\small{~=~\big[\frac{-\pi}{6}~+~\left(\frac{2}{3} \right)\left[\tan^{-1}(2 \tan \frac{\pi}{2})  \right]\big]~-~\big[0~+~\left(\frac{2}{3} \right)\left[\tan^{-1}(2 \tan 0)  \right]\big]}$

$\small{~=~\big[\frac{-\pi}{6}~+~\left(\frac{2}{3} \right)\left[\tan^{-1}(\infty)  \right]\big]~-~\big[0~+~\left(\frac{2}{3} \right)\left[\tan^{-1}(0)  \right]\big]}$

$\small{~=~\big[\frac{-\pi}{6}~+~\left(\frac{2}{3} \right)\left[\frac{\pi}{2}  \right]\big]~-~\big[0~+~\left(\frac{2}{3} \right)\left[0  \right]\big]}$

$\small{~=~\big[\frac{-\pi}{6}~+~\frac{2\pi}{6}\big]~=~\frac{\pi}{6}}$

Solved Example 23.157
Evaluate $\small{\int_{0}^{1}{\left[\frac{1}{\sqrt{1+x}~-~\sqrt{x}} \right]dx}}$
Solution:
1. Let us rearrange the given expression:

$\small{\frac{1}{\sqrt{1+x}~-~\sqrt{x}}~=~\frac{1(\sqrt{1+x}~+~\sqrt{x})}{(\sqrt{1+x}~-~\sqrt{x})(\sqrt{1+x}~+~\sqrt{x})}}$

$\small{~=~\frac{\sqrt{1+x}~+~\sqrt{x}}{1+x~-~x}~=~\sqrt{1+x}~+~\sqrt{x}}$

2. So we can write:

$\small{\int_{0}^{1}{\left[\frac{1}{\sqrt{1+x}~-~\sqrt{x}} \right]dx}~=~\int_{0}^{1}{\left[\sqrt{1+x}~+~\sqrt{x} \right]dx}}$

3. Now we can write the indefinite integral:

$\small{F~=~\int{\left[\sqrt{1+x}~+~\sqrt{x} \right]dx}~=~\frac{(1+x)^{3/2}}{3/2}~+~\frac{(x)^{3/2}}{3/2}}$

$\small{~=~\frac{(1+x)^{3/2}~+~(x)^{3/2}}{3/2}}$

4. Now we can evaluate the definite integral. We get:

$\small{I~=~\int_{0}^{1}{\left[\sqrt{1+x}~+~\sqrt{x} \right]dx}}$

$\small{~=~\big[\frac{(1+x)^{3/2}~+~(x)^{3/2}}{3/2}\big]_{0}^{1}}$

$\small{~=~\big[\frac{(1+1)^{3/2}~+~(1)^{3/2}}{3/2}\big]~-~\big[\frac{(1+0)^{3/2}~+~(0)^{3/2}}{3/2}\big]}$

$\small{~=~\big[\frac{2^{3/2}~+~1}{3/2}\big]~-~\big[\frac{1~+~0}{3/2}\big]~=~\frac{2^{3/2}}{3/2}~=~\frac{2^{5/2}}{3}~=~\frac{4 \sqrt2}{3}}$

Solved Example 23.158
Evaluate $\small{\int_{0}^{\pi}{\left[\frac{x \tan x}{\sec x~+~\tan x} \right]dx}}$
Solution:
1. Let us rearrange the given expression:

$\small{\frac{x \tan x}{\sec x~+~\tan x}~=~\frac{x \frac{\sin x}{\cos x}}{\frac{1}{\cos x}~+~\frac{\sin x}{\cos x}}~=~\frac{x \sin x}{1~+~\sin x}}$

2. So we want:

$\small{I~=~\int_{0}^{\pi}{\left[\frac{x \tan x}{\sec x~+~\tan x} \right]dx}~=~\int_{0}^{\pi}{\left[\frac{x \sin x}{1~+~\sin x} \right]dx}}$

3. Applying P₄, we get:

$\small{I~=~\int_{0}^{\pi}{\left[\frac{(\pi-x) \sin (\pi-x)}{1~+~\sin (\pi-x)} \right]dx}~=~\int_{0}^{\pi}{\left[\frac{(\pi-x) \sin x}{1~+~\sin x} \right]dx}}$

$\small{~=~\int_{0}^{\pi}{\left[\frac{\pi \sin x}{1~+~\sin x} \right]dx}~-~\int_{0}^{\pi}{\left[\frac{ x \sin x}{1~+~\sin x} \right]dx}}$

$\small{\Rightarrow I~=~\int_{0}^{\pi}{\left[\frac{\pi \sin x}{1~+~\sin x} \right]dx}~-~I}$

$\small{\Rightarrow 2I~=~\int_{0}^{\pi}{\left[\frac{\pi \sin x}{1~+~\sin x} \right]dx}}$

$\small{\Rightarrow I~=~\frac{\pi}{2}\int_{0}^{\pi}{\left[\frac{ \sin x}{1~+~\sin x} \right]dx}}$

$\small{\Rightarrow I~=~\frac{\pi}{2}\int_{0}^{\pi}{\left[\frac{\sin x (1~-~\sin x)}{(1~+~\sin x)(1~-~\sin x)} \right]dx}}$

$\small{~=~\frac{\pi}{2} \int_{0}^{\pi}{\left[\frac{\sin x~-~\sin^2 x}{1~-~\sin^2 x} \right]dx}~=~\frac{\pi}{2} \int_{0}^{\pi}{\left[\frac{\sin x~-~\sin^2 x}{\cos^2 x} \right]dx}}$

$\small{~=~\frac{\pi}{2} \int_{0}^{\pi}{\left[\sec x\, \tan x~-~\tan^2 x \right]dx}}$

$\small{~=~\frac{\pi}{2} \int_{0}^{\pi}{\left[\sec x\, \tan x~-~\left(\sec^2 x - 1 \right) \right]dx}}$

$\small{\Rightarrow I~=~\frac{\pi}{2} \int_{0}^{\pi}{\left[\sec x\, \tan x \right]dx}~-~\frac{\pi}{2} \int_{0}^{\pi}{\left[\sec^2 x \right]dx}~+~\frac{\pi}{2} \int_{0}^{\pi}{\left[1 \right]dx}}$

$\small{\Rightarrow I~=~I_1~-~I_2~+~I_3}$

4. First we will calculate I₁:

$\small{F_1~=~\frac{\pi}{2} \int{\left[\sec x\, \tan x \right]dx}~=~\frac{\pi}{2}\,\sec x}$

$\small{\Rightarrow I_1~=~\left[\frac{\pi}{2}\,\sec x \right]_{0}^{\pi}~=~\left[\frac{\pi}{2}\,\sec \left(\pi \right) \right]~-~\left[\frac{\pi}{2}\,\sec \left(0 \right) \right]}$

$\small{\Rightarrow I_1~=~\left[\frac{-\pi}{2} \right]~-~\left[\frac{\pi}{2} \right]~=~-\pi}$

5. Next we will calculate I₂:

$\small{F_2~=~\frac{\pi}{2} \int{\left[\sec^2 x \right]dx}~=~\frac{\pi}{2}\,\tan x}$

$\small{\Rightarrow I_2~=~\left[\frac{\pi}{2}\,\tan x \right]_{0}^{\pi}~=~\left[\frac{\pi}{2}\,\tan \left(\pi \right) \right]~-~\left[\frac{\pi}{2}\,\tan \left(0 \right) \right]}$

$\small{\Rightarrow I_2~=~\left[0 \right]~-~\left[0 \right]~=~0}$

6. Finally we will calculate I₃:

$\small{F_3~=~\frac{\pi}{2} \int{\left[1\right]dx}~=~\frac{\pi}{2}\,x}$

$\small{\Rightarrow I_3~=~\left[\frac{\pi}{2}\,(x) \right]_{0}^{\pi}~=~\left[\frac{\pi}{2}\, \left(\pi \right) \right]~-~\left[\frac{\pi}{2} \left(0 \right) \right]~=~\frac{\pi^2}{2}}$

7. From (3), (4), (5) and (6), we get:

$\small{I~=~I_1~-~I_2~+~I_3}$

$\small{~=~-\pi~-~0~+~\frac{\pi^2}{2}}$

$\small{~=~\left(\frac{\pi^2}{2}~-~\pi \right)~=~\left(\frac{\pi^2}{2}~-~\frac{2 \pi}{2} \right)}$

$\small{~=~\frac{\pi}{2}\left(\pi~-~2 \right)}$

Solved Example 23.159
Evaluate $\small{\int_{0}^{\frac{\pi}{4}}{\left[\frac{\sin x~+~\cos x}{9~+~16 \sin(2x)} \right]dx}}$
Solution:
1. Put $\small{u = \sin x ~-~\cos x}$

• Then $\small{\frac{du}{dx}~=~\cos x ~+~\sin x}$

$\small{\Rightarrow dx(\cos x ~+~\sin x)~=~du}$

• Also, $\small{u^2~=~\sin^2 x~+~\cos^2 x~-~2 \sin x \cos x~=~1~-~2 \sin x \cos x}$

$\small{\Rightarrow u^2~=~1~-~\sin(2x)}$

$\small{\Rightarrow \sin(2x)~=~1~-~u^2}$

2. So we want:

$\small{F ~=~\int{\left[\frac{\sin x~+~\cos x}{9~+~16 \sin(2x)} \right]dx}~=~\int{\left[\frac{1}{9~+~16(1-u^2)} \right]du}}$

$\small{~=~\int{\left[\frac{1}{9~+~16-16u^2} \right]du}~=~\int{\left[\frac{1}{25-16u^2} \right]du}~=~\int{\left[\frac{1}{5^2-(4u)^2} \right]du}}$

3. This is a standard integral. We have:

$\small{\int{\left[\frac{1}{a^2~-~t^2} \right]dt}~=~\frac{1}{2a} \log \left| \frac{a+t}{a-t}  \right|~+~\rm{C}}$

In our present case, a = 5 and t = 4u

• So we get:

$\small{F~=~\int{\left[\frac{1}{5^2-(4u)^2} \right]du}~=~\frac{1}{2(5)(4)} \log \left| \frac{5+4u}{5-4u}  \right|~=~\frac{1}{40} \log \left| \frac{5+4u}{5-4u}  \right|}$

4. Substituting for u, we get:

$\small{F~=~\frac{1}{40} \log \left|\frac{5~+~4 \sin x~-~4 \cos x}{5~-~4 \sin x~+~4 \cos x}  \right|}$

5. Now we can find the definite integral:

$\small{I~=~\left[\frac{1}{40} \log \left|\frac{5~+~4 \sin x~-~4 \cos x}{5~-~4 \sin x~+~4 \cos x}  \right| \right]_0^{\frac{\pi}{4}}}$

$\small{~=~\left[\frac{1}{40} \log \left|\frac{5~+~4 \sin \left(\frac{\pi}{4} \right)~-~4 \cos \left(\frac{\pi}{4} \right)}{5~-~4 \sin \left(\frac{\pi}{4} \right)~+~4 \cos \left(\frac{\pi}{4} \right)}  \right| \right]~-~\left[\frac{1}{40} \log \left|\frac{5~+~4 \sin \left(0 \right)~-~4 \cos \left(0 \right)}{5~-~4 \sin \left(0 \right)~+~4 \cos \left(0 \right)}  \right| \right]}$

$\small{~=~\left[\frac{1}{40} \log \left|\frac{5~+~0}{5~-~0}  \right| \right]~-~\left[\frac{1}{40} \log \left|\frac{5~+~0~-~4}{5~-~0~+~4}  \right| \right]}$

$\small{~=~\left[\frac{1}{40} \log \left|1  \right| \right]~-~\left[\frac{1}{40} \log \left|\frac{1}{9}  \right| \right]}$

$\small{~=~\left[\frac{1}{40} (0) \right]~-~\left[\frac{1}{40} \log \left|\frac{1}{9}  \right| \right]~=~~-~\left[\frac{1}{40} \log \left|\frac{1}{9}  \right| \right]}$

$\small{~=~\frac{1}{40} \log \left(9 \right)}$

Solved Example 23.160
Evaluate $\small{\int_{0}^{\frac{\pi}{2}}{\left[\sin(2x) \tan^{-1}(\sin x) \right]dx}}$
Solution:
1. Put $\small{u = \sin x}$

• Then $\small{\frac{du}{dx}~=~\cos x \Rightarrow dx~=~\frac{du}{\cos x}}$

• We wrote: $\small{u = \sin x}$

    ♦ When x approaches 0, u approaches 0
    ♦ When x approaches $\small{\frac{\pi}{2}}$, u approaches 1

2. We want:

$\small{I = \int_{0}^{\frac{\pi}{2}}{\left[\sin(2x) \tan^{-1}(\sin x) \right]dx}}$

$\small{= \int_{0}^{\frac{\pi}{2}}{\left[2 \sin(x)\cos(x) \tan^{-1}(\sin x) \right]dx}}$

$\small{= \int_{0}^{\frac{\pi}{2}}{\left[2 u\cos(x) \tan^{-1}(u) \right]\frac{du}{\cos x}}}$

$\small{= \int_{0}^{1}{\left[2 u \tan^{-1}(u) \right]du}= 2 \int_{0}^{1}{\left[u \tan^{-1}(u) \right]du}}$

3. We will apply the method of integration by parts

(a) Assigning first and second functions:

   ♦ Let first function be: $\small{f(u)=\tan^{-1}u}$

   ♦ Let second function be: $\small{g(u)=u}$

(b) Finding A:

$\small{A~=~\int{\left[g(u) \right]dx}~=~\frac{u^2}{2}}$

(c) $\small{\big[f(u) \left(A \right) \big]~=~\big[ \tan^{-1}u\big] \,\big[\frac{u^2}{2} \big]~=~\big[\frac{u^2\,\tan^{-1}u}{2} \big]}$

• This is the first term.

(d) $\small{f'(u)~=~\frac{1}{1 + u^2}}$

(e) $\small{\int{\big[f'(u)\,\left(A \right)  \big]du}~=~\int{\big[\frac{1}{1 + u^2}\,\left(\frac{u^2}{2} \right)\big]du}}$

$\small{~=~\int{\big[\frac{u^2}{2(1 + u^2)}\big]du}~=~\frac{1}{2}\int{\big[\frac{u^2}{1 + u^2}\big]du}}$

$\small{~=~\frac{1}{2}\int{\big[\frac{u^2 + 1 - 1}{1 + u^2}\big]du}~=~\frac{1}{2}\int{\big[1~-~\frac{1}{1 + u^2}\big]du}}$

$\small{~=~\frac{1}{2}\int{\big[1\big]du}~-~\frac{1}{2}\int{\big[\frac{1}{1 + u^2}\big]du}}$

$\small{~=~\frac{u}{2}~-~\frac{\tan^{-1}u}{2}}$

• This is the second term.

(f) So we get:

$\small{\int_{0}^{\frac{1}{2}}{\left[u \tan^{-1}(u) \right]du}~=~\text{First term - Second term}}$

$\small{~=~\big[\frac{u^2\,\tan^{-1}u}{2}\big]~-~\big[\frac{u}{2}~-~\frac{\tan^{-1}u}{2} \big]}$

$\small{~=~\frac{u^2\,\tan^{-1}u}{2}~-~\frac{u}{2}~+~\frac{\tan^{-1}u}{2}}$

$\small{~=~\frac{(u^2 + 1)(\tan^{-1}u)~-~u}{2}}$

4. So from (2), we get:

$\small{I= 2 \int_{0}^{1}{\left[u \tan^{-1}(u) \right]du}~=~2 \big[\frac{(u^2 + 1)(\tan^{-1}u)~-~u}{2} \big]_0^1}$

$\small{~=~ \big[(u^2 + 1)(\tan^{-1}u)~-~u \big]_0^1}$

$\small{~=~ \big[(1^2 + 1)(\tan^{-1}1)~-~1 \big]~-~\big[(0 + 1)(\tan^{-1}0)~-~0 \big]}$

$\small{~=~ \big[(2)(\pi/4)~-~1 \big]~-~\big[(1)(0)~-~0 \big]}$

$\small{~=~ \frac{\pi}{2}~-~1}$

---

The link below gives a few more miscellaneous examples:

Miscellaneous Exercise

---

In the next chapter, we will see applications of integrals.

Previous

Contents

Next 

Copyright©2025 Higher secondary mathematics.blogspot.com

23.30 - Miscellaneous Examples (2) on Integrals

In the previous section, we saw some miscellaneous examples on integrals. In this section, we will see a few more miscellaneous examples.

Solved Example 23.133
Integrate $\small{\frac{1}{\sqrt{x+a}~+~\sqrt{x+b}}}$    
Solution:
1. First we will rearrange the given expression:
$\small{\frac{1}{\sqrt{x+a}~+~\sqrt{x+b}}
~=~\frac{1}{\sqrt{x+a}~+~\sqrt{x+b}}\left(\frac{\sqrt{x+a}~-~\sqrt{x+b}}{\sqrt{x+a}~-~\sqrt{x+b}} \right)}$

$\small{~=~\frac{\sqrt{x+a}~-~\sqrt{x+b}}{(x+a)~-~(x+b)}~=~\frac{\sqrt{x+a}~-~\sqrt{x+b}}{a-b}}$

2. So we want: $\small{I = \int{\left[\frac{\sqrt{x+a}~-~\sqrt{x+b}}{a-b}\right]dx}}$

$\small{~=~\frac{1}{a-b}\int{\left[\sqrt{x+a}~-~\sqrt{x+b}\right]dx}}$

$\small{~=~\frac{1}{a-b}\int{\left[\sqrt{x+a}\right]dx}~-~\frac{1}{a-b}\int{\left[\sqrt{x+b}\right]dx}}$

3. This integration gives:

$\small{\frac{1}{a-b}\left[\frac{(x+a)^{3/2}}{3/2} \right]~-~\frac{1}{a-b}\left[\frac{(x+b)^{3/2}}{3/2} \right]}$

$\small{~=~\frac{2}{3(a-b)}\left[(x+a)^{3/2}~-~(x+b)^{3/2} \right]}$

Solved Example 23.134
Integrate $\small{\sqrt{\frac{1-\sqrt x}{1+\sqrt x}}}$
Solution:
1. First we will rearrange the given expression:

Let $\small{\sqrt x = \cos(2 \theta)}$. Then we get:

$\small{\sqrt{\frac{1-\sqrt x}{1+\sqrt x}}
~=~\sqrt{\frac{1-\cos(2 \theta)}{1+\cos(2 \theta)}}~=~\sqrt{\tan^2(\theta)}~=~\tan \theta}$

Also, since $\small{\sqrt x = \cos(2 \theta)}$, we can write:

$\small{\frac{1}{2 \sqrt x} \frac{dx}{d\theta}~=~-2\sin (2\theta)}$

$\small{\Rightarrow dx~=~-4 \sqrt x\, \sin(2 \theta) d \theta}$

$\small{\Rightarrow dx~=~-4 \cos(2\theta)\, \sin(2 \theta) d \theta}$

2. So we want: $\small{I = \int{\left[\sqrt{\frac{1-\sqrt x}{1+\sqrt x}}\right]dx}= \int{\left[\tan \theta\right]dx}}$

$\small{~=~\int{\left[\tan \theta\right][-4 \cos(2\theta)\, \sin(2 \theta) d \theta]}}$

$\small{~=~(-4)\int{\left[\tan \theta \,\cos(2\theta)\, \sin(2 \theta)\right]d \theta}}$

$\small{~=~(-4)\int{\left[\frac{\sin \theta}{\cos \theta} \,\cos(2\theta)\, 2 \sin \theta \cos \theta\right]d \theta}}$

$\small{~=~(-4)\int{\left[2 \sin^2 \theta \,\cos(2\theta) \right]d \theta}}$

$\small{~=~(-4)\int{\left[[1- \cos(2\theta)] \,\cos(2\theta) \right]d \theta}}$

$\small{~=~(-4)\int{\left[\cos(2\theta)- \cos^2(2\theta) \right]d \theta}}$

$\small{~=~(-4)\int{\left[\cos(2\theta)- \left(\frac{1+\cos(4 \theta)}{2} \right) \right]d \theta}}$

$\small{~=~(-4)\int{\left[\cos(2\theta) \right]d \theta}~-~(-4)\int{\left[\frac{1}{2}  \right]d \theta}~-~(-4)\int{\left[\frac{\cos(4 \theta)}{2}  \right]d \theta}}$

$\small{~=~(-4)\int{\left[\cos(2\theta) \right]d \theta}~+~(2)\int{\left[1  \right]d \theta}~+~(2)\int{\left[\cos(4 \theta) \right]d \theta}}$

3. This integration gives:

$\small{I~=~(-4)\left[\frac{\sin(2\theta)}{2} \right]~+~(2)\left[\theta  \right]~+~(2)\left[\frac{\sin(4\theta)}{4} \right]}$

$\small{~=~(-2)\left[\sin(2\theta) \right]~+~2 \theta~+~\frac{\sin(4\theta)}{2}}$

$\small{~=~(-2)\left[\sin(2\theta) \right]~+~2 \theta~+~\sin(2 \theta) \cos (2 \theta)}$

4. since $\small{\sqrt x = \cos(2 \theta)}$, we can write:

$\small{2 \theta~=~\cos^{-1}\left(\sqrt x \right)}$

$\small{\Rightarrow \theta~=~\frac{\cos^{-1}\left(\sqrt x \right)}{2}}$

$\small{\Rightarrow \sin (2\theta)~=~\sqrt {1 - \cos^2(2\theta)}~=~\sqrt{1-x}}$

5. Substituting the above results in (3), we get:

$\small{I~=~(-2)\left[\sin(2\theta) \right]~+~2 \theta~+~\sin(2 \theta) \cos (2 \theta)}$

$\small{~=~(-2)\left[\sqrt{1-x}\right]~+~\cos^{-1}\left(\sqrt x \right)~+~\left[\sqrt{1-x}\, \sqrt{x} \right]}$

$\small{~=~(-2)\left[\sqrt{1-x}\right]~+~\cos^{-1}\left(\sqrt x \right)~+~\left[\sqrt{x-x^2} \right]~+~\rm{C}}$

Solved Example 23.135
Integrate $\small{\frac{\sin^{-1} \sqrt{x} - \cos^{-1} \sqrt{x}}{\sin^{-1} \sqrt{x} + \cos^{-1} \sqrt{x}},~x \in [0,1]}$
Solution:
1. First we will rearrange the given expression:

Let $\small{\sin^{-1} \sqrt x = \theta}$. Then we get:

$\small{\sin \theta~=~\sqrt x}$

$\small{\Rightarrow \cos\left(\frac{\pi}{2} - \theta\right)~=~\sqrt x}$

$\small{\Rightarrow \cos^{-1} \sqrt x ~=~\frac{\pi}{2} - \theta}$

• Also, since $\small{\sin \theta~=~\sqrt x}$, we can write:

$\small{\cos \theta \frac{d \theta}{dx}~=~\frac{1}{2 \sqrt x}}$

$\small{\Rightarrow dx~=~2 \sqrt x \cos \theta \,d\theta}$

$\small{\Rightarrow dx~=~2 \sin\theta \cos \theta \,d\theta}$

$\small{\Rightarrow dx~=~\sin(2\theta) \,d\theta}$

2. So we want: $\small{I = \int{\left[\frac{\sin^{-1} \sqrt{x} ~-~ \cos^{-1} \sqrt{x}}{\sin^{-1} \sqrt{x} ~+~ \cos^{-1} \sqrt{x}}\right]dx}}$

$\small{~=~\int{\left[\frac{\theta~-~\left(\frac{\pi}{2} - \theta\right)}{\theta~+~\left(\frac{\pi}{2} - \theta\right)} \right][\sin(2\theta) \,d\theta]}}$

$\small{~=~\int{\left[\frac{2\theta~-~\frac{\pi}{2}}{\frac{\pi}{2}} \right][\sin(2\theta) \,d\theta]}}$

$\small{~=~\int{\left[\frac{4 \theta}{\pi} ~-~1\right][\sin(2\theta) \,d\theta]}}$

$\small{~=~\int{\left[\frac{4 \theta}{\pi}\right][\sin(2\theta) \,d\theta]}~-~\int{\left[1\right][\sin(2\theta) \,d\theta]}}$

$\small{~=~\int{\left[\frac{4 \theta}{\pi} \sin(2\theta)\right]d\theta}~-~\int{\left[\sin(2\theta)\right]d\theta}}$

$\small{~=~\frac{4}{\pi}\int{\left[\theta\, \sin(2\theta)\right]d\theta}~-~\int{\left[\sin(2\theta)\right]d\theta}}$

$\small{~=~\left(\frac{4}{\pi} \right)I_1~-~I_2}$

3. Next we will calculate I₁:

(a) Assigning first and second functions:

   ♦ Let first function be: $\small{f(\theta)=\theta}$

   ♦ Let second function be: $\small{g(\theta)=\sin(2\theta)}$

(b) Finding A:

$\small{A~=~\int{\left[g(\theta) \right]dx}~=~\int{\left[\sin(2\theta)\right]d\theta}~=~\frac{- \cos (2 \theta)}{2}}$

(c) $\small{\big[f(\theta) \left(A \right) \big]~=~\big[\frac{-\theta \cos (2 \theta)}{2} \big]}$

• This is the first term.

(d) $\small{f'(\theta)~=~1}$

(e) $\small{\int{\big[f'(\theta)\,\left(A \right)  \big]d\theta}~=~\int{\big[1\,\left(\frac{- \cos (2 \theta)}{2} \right)\big]d\theta}}$

$\small{~=~\frac{-1}{2}\int{\big[\cos(2\theta)\big]d\theta}~=~\frac{-\sin(2\theta)}{4}}$

• This is the second term.

(f) So we get:

$\small{I_1~=~\text{First term - Second term}}$

$\small{~=~\big[\frac{-\theta \cos (2 \theta)}{2}\big]~-~\big[\frac{-\sin(2\theta)}{4}\big]~=~\frac{\sin(2\theta)~-~2 \theta \cos(2 \theta)}{4}}$

4. Next we calculate I₂:

$\small{\int{\left[\sin(2\theta)\right]d\theta}~=~\frac{-\cos(2\theta)}{2}}$

5. So from (2), we get: $\small{I = \left(\frac{4}{\pi} \right)I_1 - I_2}$

$\small{~=~\left(\frac{4}{\pi} \right)\frac{\sin(2\theta)~-~2 \theta \cos(2 \theta)}{4}~-~\frac{-\cos(2\theta)}{2}}$

$\small{~=~\frac{\sin(2\theta)~-~2 \theta \cos(2 \theta)}{\pi}~+~\frac{\cos(2\theta)}{2}}$

6. Now we can substitute for $\small{\theta}$.

• Since $\small{\sin \theta~=~\sqrt x}$, we can write:

$\small{\cos \theta ~=~\sqrt{1-x}}$

$\small{\Rightarrow \sin(2 \theta) ~=~2 \sin \theta\,\cos \theta~=~2\sqrt{x(1-x)}~=~2\sqrt{(x-x^2)}}$

$\small{\Rightarrow \cos(2 \theta) ~=~\cos^2\theta~-~\sin^2 \theta~=~1-2x}$

7. Based on the result in (5), we get:

$\small{I~=~\frac{\sin(2\theta)~-~2 \theta \cos(2 \theta)}{\pi}~+~\frac{\cos(2\theta)}{2}}$

$\small{~=~\frac{2\sqrt{(x-x^2)}~-~\left[2 \sin^{-1}\sqrt{x} \right] (1-2x)}{\pi}~+~\frac{1-2x}{2}}$

$\small{~=~\frac{2\sqrt{(x-x^2)}~+~\left[2 \sin^{-1}\sqrt{x} \right] (2x-1)}{\pi}~+~\frac{1-2x}{2}}$

$\small{~=~\frac{2\sqrt{(x-x^2)}~+~\left[2 \sin^{-1}\sqrt{x} \right] (2x-1)}{\pi}~+~\frac{1}{2}~-~x~+~\rm{C_1}}$

$\small{~=~\frac{2\sqrt{(x-x^2)}~+~\left[2 \sin^{-1}\sqrt{x} \right] (2x-1)}{\pi}~-~x~+~\rm{C}}$

Solved Example 23.136
Integrate $\small{\frac{\sin x}{\sin(x-a)}}$
Solution:
1. First we will rearrange the given expression:

Let $\small{u = x-a}$. Then we get:

$\small{x~=~u+a}$

Also, $\small{\frac{du}{dx}~=~1 \Rightarrow du = dx}$

2. So we want: $\small{I = \int{\left[\frac{\sin x}{\sin(x-a)}\right]dx} = \int{\left[\frac{\sin (u+a)}{\sin(u)}\right]du}}$

$\small{= \int{\left[\frac{\sin u \cos a~+~\cos u \sin a}{\sin u}\right]du}= \int{\left[\cos a~+~\cot u \sin a\right]du}}$

$\small{= \int{\left[\cos a\right]du}~+~\int{\left[\cot u \sin a\right]du}}$

$\small{= \cos a \int{\left[1\right]du}~+~\sin a \int{\left[\cot u\right]du}}$

$\small{= I_1~+~I_2}$

3. Next we calculate I₁. We get:

$\small{= \cos a \int{\left[1\right]du}~=~\cos a \,(u)}$

4. Next we calculate I₂:

$\small{\sin a \int{\left[\cot u\right]du}~=~ (\sin a)\log \left|\sin u \right|}$

5. So we can write:

$\small{I=I_1 + I_2~=~\cos a \,(u)~+~(\sin a)\log \left|\sin u \right|}$

• Substituting for u, we get:

$\small{I~=~\cos a \,(x-a)~+~(\sin a)\log \left|\sin (x-a) \right|~+~\rm{C_1}}$

$\small{~=~x \cos a ~-~ a \cos a~+~(\sin a)\log \left|\sin (x-a) \right|~+~\rm{C_1}}$

$\small{~=~x \cos a~+~(\sin a)\log \left|\sin (x-a) \right|~+~\rm{C}}$

Solved Example 23.137
Integrate $\small{\frac{\sin^8 x~-~\cos^8 x}{1~-~2 \sin^2 x \cos^2 x}}$
Solution:
1. First we will rearrange the given expression:

$\small{\frac{(\sin^4 x)^2~-~(\cos^4 x)^2}{1~-~2 \sin^2 x \cos^2 x}~=~\frac{\left[(\sin^4 x)~+~(\cos^4 x)\right] \left[(\sin^4 x)~-~(\cos^4 x)\right]}{1~-~2 \sin^2 x \cos^2 x}}$

$\small{~=~\frac{\left[(\sin^2 x)^2~+~(\cos^2 x)^2 ~+~2 \sin^2 x \cos^2 x  ~-~2 \sin^2 x \cos^2 x\right] \left[(\sin^4 x)~-~(\cos^4 x)\right]}{1~-~2 \sin^2 x \cos^2 x}}$

$\small{~=~\frac{\left[[(\sin^2 x)~+~(\cos^2 x)]^2  ~-~2 \sin^2 x \cos^2 x\right] \left[(\sin^4 x)~-~(\cos^4 x)\right]}{1~-~2 \sin^2 x \cos^2 x}}$

$\small{~=~\frac{\left[[1]^2  ~-~2 \sin^2 x \cos^2 x\right] \left[(\sin^4 x)~-~(\cos^4 x)\right]}{1~-~2 \sin^2 x \cos^2 x}}$

$\small{~=~\left[(\sin^4 x)~-~(\cos^4 x)\right]}$

$\small{~=~\left[(\sin^2 x)~+~(\cos^2 x)\right]~\left[(\sin^2 x)~-~(\cos^2 x)\right]}$

$\small{~=~\left[1\right]~\left[(\sin^2 x)~-~(\cos^2 x)\right]}$

$\small{~=~\left[-1\right]~\left[(\cos^2 x)~-~(\sin^2 x)\right]}$

$\small{~=~- \cos(2x)}$

2. So we get: $\small{I = \int{\left[-\cos(2x)\right]dx} = \frac{-\sin(2x)}{2}~+~\rm{C}}$

Solved Example 23.138
Integrate $\small{\frac{1}{\cos (x+a)\, \cos(x+b)}}$
Solution:
1. First we will rearrange the given expression:

$\small{\frac{1}{\cos (x+a)\, \cos(x+b)}~=~\frac{\sin(a-b)}{\sin(a-b)}\times\frac{1}{\cos (x+a)\, \cos(x+b)}}$

$\small{~=~\frac{1}{\sin(a-b)}\times\frac{\sin(a-b)}{\cos (x+a)\, \cos(x+b)}~=~\frac{1}{\sin(a-b)}\times\frac{\sin(a-b-x+x)}{\cos (x+a)\, \cos(x+b)}}$

$\small{~=~\frac{1}{\sin(a-b)}\times\frac{\sin\left[ (x+a)-(x+b)\right]}{\cos (x+a)\, \cos(x+b)}}$

$\small{~=~\frac{1}{\sin(a-b)}\times\frac{\sin(x+a) \cos(x+b)~-~\cos(x+a) \sin(x+b)}{\cos (x+a)\, \cos(x+b)}}$

$\small{~=~\frac{1}{\sin(a-b)}\times\left[\frac{\sin(x+a) \cos(x+b)}{\cos (x+a)\, \cos(x+b)}~-~\frac{\cos(x+a) \sin(x+b)}{\cos (x+a)\, \cos(x+b)} \right]}$

$\small{~=~\frac{1}{\sin(a-b)}\times\left[\frac{\sin(x+a)}{\cos (x+a)}~-~\frac{ \sin(x+b)}{\cos(x+b)} \right]}$

$\small{~=~\frac{1}{\sin(a-b)}\times\left[\tan(x+a)~-~\tan(x+b) \right]}$

2. So we get:

$\small{I = \int{\left[\frac{1}{\cos (x+a)\, \cos(x+b)}\right]dx} = \int{\left[\frac{1}{\sin(a-b)}\times\left[\tan(x+a)~-~\tan(x+b) \right]\right]dx}}$

$\small{= \frac{1}{\sin(a-b)} \int{\left[\tan(x+a)~-~\tan(x+b) \right]dx}}$

$\small{= \frac{1}{\sin(a-b)} \left[\log \left|\sec(x+a) \right|~-~\log \left|\sec(x+b) \right| \right]~+~\rm{C}}$

$\small{= \frac{1}{\sin(a-b)} \left[\log \left| \frac{\sec(x+a)}{\sec(x+b)}  \right| \right]~+~\rm{C}}$

$\small{= \frac{1}{\sin(a-b)} \left[\log \left| \frac{\cos(x+b)}{\cos(x+a)}  \right| \right]~+~\rm{C}}$

Solved Example 23.139
Integrate $\small{\frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}}}$
Solution:
1. First we will rearrange the portion inside square root symbol:

$\small{\sin^3 x \sin(x+\alpha)}$

$\small{~=~\sin^3 x \left[\sin x \cos \alpha~+~\cos x \sin\alpha \right]}$

$\small{~=~\sin^4 x \cos \alpha~+~\sin^3 x\cos x \sin\alpha }$

$\small{~=~\sin^4 x \cos \alpha~+~\sin^3 x\cos x \sin\alpha \frac{\sin x}{\sin x}}$

$\small{~=~\sin^4 x \cos \alpha~+~\sin^4 x\cot x \sin\alpha}$

$\small{~=~\sin^4 x \left[\cos \alpha~+~\cot x \sin\alpha \right]}$

2. So we get:

$\small{I = \int{\left[\frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}}\right]dx} =\int{\left[\frac{1}{\sqrt{\sin^4 x \left[\cos \alpha~+~\cot x \sin\alpha \right]}}\right]dx}}$

$\small{=\int{\left[\frac{1}{\sin^2 x \sqrt{\cos \alpha~+~\cot x \sin\alpha }}\right]dx}=\int{\left[\frac{\csc^2 x}{\sqrt{\cos \alpha~+~\cot x \sin\alpha }}\right]dx}}$

3. Let us put $\small{u = \cos \alpha~+~\cot x\, \sin\alpha}$

• Then $\small{\frac{du}{dx}~=~0 + \sin\alpha(-\csc^2 x)~=~-\sin \alpha\,\csc^2 x}$

$\small{\Rightarrow -\sin \alpha\,\csc^2 x\,dx~=~du}$

4. So we want:

$\small{I=\int{\left[\frac{\csc^2 x}{\sqrt{\cos \alpha~+~\cot x \sin\alpha }}\right]dx}=\int{\left[\frac{(-\sin \alpha)\csc^2 x}{(-\sin \alpha)\sqrt{\cos \alpha~+~\cot x \sin\alpha }}\right]dx}}$

$\small{=\int{\left[\frac{1}{(-\sin \alpha)\sqrt{u}}\right]du}=\frac{1}{(-\sin \alpha)}\int{\left[\frac{1}{\sqrt{u}}\right]du}}$

5. This integration gives:

$\small{\frac{1}{(-\sin \alpha)}\left[\frac{u^{1/2}}{1/2}~+~\rm{C_1}\right]~=~\frac{-2\,u^{1/2}}{\sin\alpha}~+~\rm{C}}$

6. Substituting for u, we get:

$\small{I~=~\frac{-2\,(\cos \alpha~+~\cot x\, \sin\alpha)^{1/2}}{\sin\alpha}~+~\rm{C}}$

7. From step (1), we have:

$\small{\sin^3 x \sin(x+\alpha)~=~\sin^4 x \left[\cos \alpha~+~\cot x \sin\alpha \right]}$

$\small{\Rightarrow \sin(x+\alpha)~=~\sin x \left[\cos \alpha~+~\cot x \sin\alpha \right]}$

$\small{\Rightarrow \cos \alpha~+~\cot x \sin\alpha~=~\frac{\sin(x+\alpha)}{\sin x}}$

• So from (6), we get:

$\small{I~=~\frac{-2}{\sin\alpha} \sqrt{\frac{\sin(x+\alpha)}{\sin x}}~+~\rm{C}}$

Solved Example 23.140
Integrate $\small{\tan^{-1}\sqrt{\frac{1-x}{1+x}}}$
Solution:
1. First we will rearrange the given expression:

Let $\small{x = \cos(2 \theta)}$. Then we get:

$\small{\sqrt{\frac{1-x}{1+ x}}
~=~\sqrt{\frac{1-\cos(2 \theta)}{1+\cos(2 \theta)}}~=~\sqrt{\tan^2(\theta)}~=~\tan \theta}$

• Also, since $\small{x = \cos(2 \theta)}$, we can write:

$\small{\frac{dx}{d\theta}~=~-2\sin (2\theta)}$

$\small{\Rightarrow dx~=~-2 \sin(2 \theta) d \theta}$

• Also, since $\small{x = \cos(2 \theta)}$, we can write:

$\small{2\theta~=~\cos^{-1}x ~~\text{and}~~ \sin(2\theta)~=~\sqrt{1 - x^2}}$

2. So we want: $\small{I = \int{\left[\tan^{-1}\sqrt{\frac{1-x}{1+x}}\right]dx}= \int{\left[\tan^{-1}(\tan \theta)\right]dx}}$

$\small{~=~ \int{\left[\theta\right]dx}~=~\int{\left[\theta\right]}(-2 \sin(2 \theta) d \theta)~=~-2\int{\left[\theta \sin(2 \theta)\right]}d \theta}$

$\small{\int{\left[\theta \sin(2 \theta)\right]}d \theta}$ is already done in solved example 23.135 above.

So we get:

$\small{I~=~-2\left[\frac{\sin(2\theta)~-~2 \theta \cos(2 \theta)}{4}~+~\rm{C_1}\right]~=~(-1)\left[\frac{\sin(2\theta)~-~2 \theta \cos(2 \theta)}{2}~+~\rm{C_2}\right]}$

$\small{~=~\frac{1}{2}\left[2 \theta \cos(2 \theta)~-~\sin(2\theta)~+~\rm{C_3}\right]}$

3. Substituting for $\small{2\theta,~ \cos(2 \theta)~\rm{and}~\sin(2 \theta),}$ we get:

$\small{I~=~\frac{1}{2}\left[\cos^{-1}x\,(x)~-~\sqrt{1 - x^2}~+~\rm{C_3}\right]}$

$\small{\Rightarrow I~=~\frac{1}{2}\left[(x) \cos^{-1}x\,~-~\sqrt{1 - x^2}\right]~+~\rm{C}}$

Solved Example 23.141
Integrate $\small{\frac{\cos(2x)}{(\sin x + \cos x)^2}}$
Solution:
1. First we will rearrange the given expression:

$\small{\frac{\cos(2x)}{(\sin x + \cos x)^2}
~=~\frac{\cos(2x)}{\sin^2 x +2 \sin x \cos x + \cos^2 x}~=~\frac{\cos(2x)}{1 +2 \sin x \cos x}~=~\frac{\cos(2x)}{1 + \sin (2x)}}$

• The derivative of [1+sin(2x)] is cos(2x).

• So we put $\small{u = 1+\sin(2x)\Rightarrow \frac{du}{dx}~=~2 \cos (2x)}$

$\small{\Rightarrow \frac{du}{dx}~=~2 \cos (2x) \Rightarrow du~=~2 \cos(2x) dx}$

2. So we want: $\small{I = \int{\left[\frac{\cos(2x)}{(\sin x + \cos x)^2}\right]dx}= \int{\left[\frac{2\cos(2x)}{2(1 + \sin (2x))}\right]dx}}$

$\small{~=~ \int{\left[\frac{1}{2(u)} \right]du}~=~\frac{1}{2} \int{\left[\frac{1}{u} \right]du}}$

3. This integration gives:

$\small{I~=~\frac{1}{2} \left[\log \left|u \right|~+~\rm{C_1}\right]~=~\frac{1}{2} \log \left|u \right|~+~\rm{C}}$

4. Substituting for u, we get:

$\small{I~=~\frac{1}{2} \log \left|1 + \sin (2x) \right|~+~\rm{C}}$

$\small{~=~\frac{1}{2} \log \left|(\sin x + \cos x)^2 \right|~+~\rm{C}}$

$\small{~=~\frac{2}{2} \log \left|(\sin x + \cos x) \right|~+~\rm{C}}$

$\small{~=~ \log \left|(\sin x + \cos x) \right|~+~\rm{C}}$

Solved Example 23.142
Integrate $\small{\frac{5x}{(x+1)(x^2 + 9)}}$
Solution:
1. The numerator is a polynomial of degree 1. The denominator is a polynomial of degree 3.

2. So it is a proper rational function. We can straight away start partial fraction decomposition

3. The denominator is already factorized:
$\small{(x+1)(x^2 + 9)}$

   ♦ One factor is quadratic.
   
   ♦ That quadratic factor cannot be further factorized.
   
   ♦ All other factors are linear.

4. The quadratic factor $\small{(x^2 + 9)}$ cannot be further factorized. Then we are able to write:

$\small{\frac{5x}{(x+1)(x^2 + 9)}~=~\left[\frac{Ax + B}{x^2 + 9}\right]~+~\frac{A_1}{x+1}}$
Where A, B and A₁ are real numbers.

5. To find A, B and A₁, we make denominators same on both sides:

$\small{\frac{5x}{(x+1)(x^2 + 9)}~=~\frac{(Ax+B) (x+1)~+~A_1 (x^2+9)}{(x+1)(x^2 + 9)}}$

6. Since denominators are same on both sides, we can equate the numerators. We get:

$\small{5x~=~(Ax+B) (x+1)~+~A_1 (x^2+9)}$

7. After equating the numerators, we can use suitable substitution.

   ♦ Put x = −1. We get: $\small{-5~=~10A_1}$. So $\small{A_1 = \frac{-1}{2}}$  
    
   ♦ Put x = 0. We get: $\small{0~=~B~-~(9/2)}$. So $\small{B = \frac{9}{2}}$  
     
   ♦ Put x = 1. We get: $\small{5~=~(A+9/2) (2)~-~5}$. So $\small{A = \frac{1}{2}}$
   
8. Now the result in (4) becomes:

$\small{\frac{5x}{(x+1)(x^2 + 9)}~=~\left[\frac{(1/2)x + 9/2}{x^2 + 9}\right]~-~\frac{1/2}{x+1}}$

$\small{~=~\left[\frac{x ~+~ 9}{2(x^2 + 9)}\right]~-~\frac{1}{2(x + 1)}}$

9. So the integration becomes easy. We get:

$\small{\frac{1}{4} \log (x^2+9)~+~\frac{3}{2} \tan^{-1}\left(\frac{x}{3} \right)~-~\frac{1}{2} \log \left|x+1 \right|~+~\rm{C}}$

• The reader may write all the steps involved in the integration process.

Solved Example 23.143
Integrate $\small{\frac{1}{(x^2 + 1)(x^2 + 4)}}$
Solution:
1. The numerator is a polynomial of degree 0. The denominator is a polynomial of degree 4.

2. So it is a proper rational function. We can straight away start partial fraction decomposition

3. The denominator is already factorized:
$\small{(x^2 + 1)(x^2 + 4)}$

   ♦ Both factors are quadratic.
   
   ♦ Both quadratic factors cannot be further factorized.
   
4. Then we are able to write:

$\small{\frac{1}{(x^2 + 1)(x^2 + 4)}~=~\left[\frac{A_1 x + B_1}{x^2 + 1}\right]~+~\left[\frac{A_2 x + B_2}{x^2 + 4}\right]}$
Where A₁, B₁, A₂ and B₂ are real numbers.

5. To find those real numbers, we make denominators same on both sides:

$\small{\frac{1}{(x^2 + 1)(x^2 + 4)}~=~\frac{(A_1 x + B_1) (x^2 + 4)~+~(A_2 x + B_2) (x^2 + 1)}{(x+1)(x^2 + 4)}}$

6. Since denominators are same on both sides, we can equate the numerators. We get:

$\small{1~=~(A_1 x + B_1) (x^2 + 4)~+~(A_2 x + B_2) (x^2 + 1)}$

7. After equating the numerators, we can use suitable substitution.

   ♦ Put x = 0. We get: $\small{1~=~4B_1~+~B_2}$
    
   ♦ Put x = −1. We get: $\small{1~=~-5A_1~+~5B_1~-~2A_2~+~2B_2}$
     
   ♦ Put x = −2. We get: $\small{1~=~-16A_1~+~8B_1~-~10A_2~+~5B_2}$
   
   ♦ Put x = 1. We get: $\small{1~=~5A_1~+~5B_1~+~2A_2~+~2B_2}$
   
• Solving the four equations, we get:

$\small{A_1 = 0,~A_2=0,~B_1=\frac{1}{3},~\rm{and}~B_2=\frac{-1}{3}}$
   
8. Now the result in (4) becomes:

$\small{\frac{1}{(x^2 + 1)(x^2 + 4)}~=~\left[\frac{1}{3(x^2 + 1)}\right]~-~\left[\frac{1}{3(x^2 + 4)}\right]}$

9. So the integration becomes easy. We get:

$\small{\frac{1}{3} \tan^{-1}\left(x \right)~-~\frac{1}{6} \tan^{-1}\left(\frac{x}{2} \right)~+~\rm{C}}$

• The reader may write all the steps involved in the integration process.

Solved Example 23.144
Integrate $\small{\frac{x^2 + x + 1}{(x + 1)^2 (x + 2)}}$
Solution:
1. The numerator is a polynomial of degree 2. The denominator is a polynomial of degree 3.

2. So it is a proper rational function. We can straight away start partial fraction decomposition

3. The denominator is already factorized:
$\small{(x + 1)^2 \,(x + 2)}$

   ♦ All factors are linear.
   
   ♦ The factor (x+1) appears twice.
   
4. Then we are able to write:

$\small{\frac{x^2 + x + 1}{(x + 1)^2 \, (x + 2)}~=~\left[\frac{A_1}{x + 1}\right]~+~\left[\frac{A_2}{(x + 1)^2}\right]~+~\left[\frac{A_3}{x + 2}\right]}$
Where A₁, A₂ and A₃ are real numbers.

5. To find those real numbers, we make denominators same on both sides:

$\small{\frac{x^2 + x + 1}{(x + 1)^2 \, (x + 2)}~=~\frac{A_1 (x + 1)(x+2)~+~A_2 (x + 2)~+~A_3 (x+1)^2}{(x + 1)^2 \, (x + 2)}}$

6. Since denominators are same on both sides, we can equate the numerators. We get:

$\small{x^2 + x + 1~=~A_1 (x + 1)(x+2)~+~A_2 (x + 2)~+~A_3 (x+1)^2}$

7. After equating the numerators, we can use suitable substitution.

   ♦ Put x = 0. We get: $\small{1~=~2 A_1~+~2 A_2~+~A_3}$
    
   ♦ Put x = −1. We get: $\small{1~=~A_2}$
     
   ♦ Put x = −2. We get: $\small{3~=~A_3}$
   
• Solving the three equations, we get:

$\small{A_1 = -2,~A_2=1,~\rm{and}~A_3=3}$
   
8. Now the result in (4) becomes:

$\small{\frac{x^2 + x + 1}{(x + 1)^2 \, (x + 2)}~=~\left[\frac{-2}{x + 1}\right]~+~\left[\frac{1}{(x + 1)^2}\right]~+~\left[\frac{3}{x + 2}\right]}$

9. So the integration becomes easy. We get:

$\small{-2 \log \left|x+1 \right|~-~\frac{1}{x+1}~+~3 \log \left|x+2 \right|~+~\rm{C}}$

• The reader may write all the steps involved in the integration process.

Solved Example 23.145
Integrate $\small{\frac{e^{5 \log x}~-~e^{4 \log x}}{e^{3 \log x}~-~e^{2 \log x}}}$
Solution:
1. Consider the first term of the numerator. Let us write:

$\small{u~=~e^{5 \log x}}$

• Taking log on both sides, we get: $\small{\log u~=~\log\left(e^{5 \log x} \right)}$

$\small{\Rightarrow \log u~=~5 \log x\left[\log\left(e \right) \right]~=~5 \log(x)~=~\log\left(x^5 \right)}$

$\small{\Rightarrow u~=~x^5}$

2. In this way, all terms can be rearranged. The given expression becomes:

$\small{\frac{x^5~-~x^4}{x^3~-~x^2}}$. This can be rearranged as: $\small{\frac{x^4 \left(x~-~1 \right)}{x^2 \left(x~-~1 \right)}~=~x^2}$

3. So we want: $\small{I = \int{\left[\frac{e^{5 \log x}~-~e^{4 \log x}}{e^{3 \log x}~-~e^{2 \log x}}\right]dx}= \int{\left[x^2 \right]dx}}$

4. This integration gives: $\small{I~=~\frac{x^3}{3}~+~\rm{C}}$

Solved Example 23.146
Integrate $\small{\cos^3 x\,e^{\log(\sin x)}}$
Solution:
1. First we will rearrange the second portion:

• Let us write:

$\small{u~=~e^{\log(\sin x)}}$

• Taking log on both sides, we get: $\small{\log u~=~\log\left(e^{\log(\sin x)} \right)}$

$\small{\Rightarrow \log u~=~\log(\sin x) \left[\log\left(e \right) \right]~=~\log(\sin x)}$

$\small{\Rightarrow u~=~\sin x}$

2. So the given expression becomes:

$\small{\cos^3 x\,\sin x}$.

3. So we want: $\small{I = \int{\left[\cos^3 x\,\sin x\right]dx}}$

4. Put $\small{t~=~\cos x}$

• Then we get:$\small{\frac{dt}{dx}~=~-\sin x \Rightarrow -\sin x \, dx~=~dt}$

5. So we want:

$\small{I = \int{\left[\cos^3 x\,\sin x\right]dx} = \int{\left[(-1)(-1)\cos^3 x\,\sin x\right]dx} = (-1)\int{\left[t^3 \right]dt}}$

6. This integration gives: $\small{I~=~(-1)\left[\frac{t^4}{4} ~+~\rm{C_1} \right]~=~(-1)\frac{t^4}{4} ~+~\rm{C}}$

7. Substituting for t, we get:

$\small{I~=~(-1)\frac{\cos^4 x}{4} ~+~\rm{C}}$

Solved Example 23.147
Integrate $\small{e^{3 \log(x)}\,(x^4 + 1)^{-1}}$
Solution:
1. First we will rearrange $\small{\left[e^{3 \log(x)} \right]}$:

• Let us write:

$\small{u~=~e^{3 \log(x)}}$

• Taking log on both sides, we get: $\small{\log u~=~\log\left(e^{3 \log(x)} \right)}$

$\small{\Rightarrow \log u~=~3 \log(x) \left[\log\left(e \right) \right]~=~3 \log(x)~=~\log\left(x^3 \right)}$

$\small{\Rightarrow u~=~x^3}$

2. So the given expression becomes:

$\small{x^3\,(x^4 + 1)^{-1}~=~\frac{x^3}{x^4 + 1}}$

3. So we want: $\small{I = \int{\left[e^{3 \log(x)}\,(x^4 + 1)^{-1}\right]dx}= \int{\left[\frac{x^3}{x^4 + 1}\right]dx}}$

4. Put $\small{t~=~x^4 + 1}$

• Then we get:$\small{\frac{dt}{dx}~=~4x^3 \Rightarrow  4x^3\, dx~=~dt}$

5. So we want:

$\small{I = \int{\left[\frac{x^3}{x^4 + 1} \right]dx} = \int{\left[\frac{(4)x^3}{(4)(x^4 + 1)} \right]dx} = \int{\left[\frac{1}{4t} \right]dt}}$

6. This integration gives: $\small{I~=~\frac{1}{4} \log \left|t \right| ~+~\rm{C}}$

7. Substituting for t, we get:

$\small{I~=~\frac{1}{4} \log \left|x^4 + 1 \right|~+~\rm{C}}$

---

In the next section, we will see a few more miscellaneous examples.

Previous

Contents

Next 

Copyright©2025 Higher secondary mathematics.blogspot.com

23.29 - Miscellaneous Examples (1) on Integrals

In the previous section, we completed a discussion on definite integrals. We saw some solved examples also. In this section, we will see some miscellaneous examples.

Solved Example 23.119
Find $\small{\int \left[\cos(6x) \sqrt{1+\sin(6x)} \right]dx}$
Solution:
1. The derivative of [1+sin(6x)] is 6cos(6x).
So we put u = 1 + sin(6x)

$\small{\Rightarrow \frac{du}{dx} = 6 \cos(6x)}$
$\small{\Rightarrow 6 \cos(6x) dx = du}$ 

2. So we want:
$\small{\int \left[\cos(6x) \sqrt{1+\sin(6x)} \right]dx~=~\int \left[\frac{6\cos(6x)}{6} \sqrt{1+\sin(6x)} \right]dx}$

$\small{~=~\int \left[\frac{1}{6} \sqrt{u} \right]dx}$

3. This integration can be done as shown below:

$\small{\int \left[\frac{1}{6} \sqrt{u} \right]dx~=~\frac{1}{6} \int \left[u^{\frac{1}{2}} \right]dx~=~\frac{1}{6} \left[\frac{u^{\frac{3}{2}}}{\frac{3}{2}} \right]+\rm{C}~=~\frac{u^{\frac{3}{2}}}{9}+\rm{C}}$

4. Substituting for u, we get:
$\small{\int \left[\cos(6x) \sqrt{1+\sin(6x)} \right]dx~=~\frac{(1+\sin(6x))^{\frac{3}{2}}}{9}+\rm{C}}$

Solved Example 23.120
Find $\small{\int \left[\frac{(x^4 - x)^{\frac{1}{4}}}{x^5} \right]dx}$
Solution:
1. First we will rearrange the given expression:
$\small{\frac{(x^4 - x)^{\frac{1}{4}}}{x^5}~=~\frac{\left[x^4 \left(1 - \frac{1}{x^3} \right) \right]^{\frac{1}{4}}}{x^5}~=~\frac{x \left[\left(1 - \frac{1}{x^3} \right) \right]^{\frac{1}{4}}}{x^5}~=~\frac{\left(1 - \frac{1}{x^3} \right)^{\frac{1}{4}}}{x^4}}$

2. The derivative of $\small{\left(1 - \frac{1}{x^3} \right)}$ is $\small{\frac{3}{x^4}}$

So we put $\small{u=1 - \frac{1}{x^3}}$

$\small{\Rightarrow \frac{du}{dx}~=~\frac{3}{x^4}~~~\Rightarrow \frac{3}{x^4}dx~=~du}$

3. So we want:

$\small{\int{\left[\frac{\left(1 - \frac{1}{x^3} \right)^{\frac{1}{4}}}{x^4} \right]dx}=\int{\left[\frac{3\left(1 - \frac{1}{x^3} \right)^{\frac{1}{4}}}{3 x^4} \right]dx}=\int{\left[\frac{\left(u \right)^{\frac{1}{4}}}{3} \right]du}}$

4. This integration can be done as shown below:

$\small{\int{\left[\frac{\left(u \right)^{\frac{1}{4}}}{3} \right]du}~=~\frac{\left(u \right)^{\frac{5}{4}}}{3(5/4)}~=~\frac{4\left(u \right)^{\frac{5}{4}}}{15}}$

4. Substituting for u, we get:

$\small{\int{\left[\frac{\left(1 - \frac{1}{x^3} \right)^{\frac{1}{4}}}{x^4} \right]dx}=\frac{4\left(1 - \frac{1}{x^3} \right)^{\frac{5}{4}}}{15}+\rm{C}=\frac{4}{15} \left(1 - \frac{1}{x^3} \right)^{\frac{5}{4}}+\rm{C}}$

Solved Example 23.121
Find $\small{\int \left[\frac{x^4}{(x-1)(x^2+1)} \right]dx}$
Solution:
1. The numerator is a polynomial of degree 4. The denominator is a polynomial of degree 3.

2. So it is not a proper rational function. We must do long division. We get:

$\small{\frac{x^4}{(x-1)(x^2+1)}\,=\,(x+1)~+~ \frac{1}{(x-1)(x^2 + 1)}}$

• The reader may write all steps involved in the long division (or any other suitable method) process.

3. In the R.H.S, the first term can be easily integrated. But the second term must be subjected to partial fraction decomposition.

• The denominator is already factorized: $\small{(x-1)(x^2 + 1)}$

   ♦ One factor is quadratic.
   
   ♦ That quadratic factor cannot be further factorized.
   
   ♦ All other factors are linear.

4. The quadratic factor $\small{(x^2+1)}$ cannot be further factorized. So this is case III. Then we are able to write:

$\small{\frac{1}{(x^2 + 1)(x - 1)}~=~\left[\frac{Ax + B}{x^2 + 1}\right]~+~\frac{A_1}{ x - 1}}$
Where A, B and A₁ are real numbers.

5. To find A, B and A₁, we make denominators same on both sides:

$\small{\frac{1}{(x^2 + 1)(x - 1)}~=~\frac{(Ax+B) (x-1)~+~A_1 (x^2+1)}{(x^2 +1)(x-1)}}$

6. Since denominators are same on both sides, we can equate the numerators. We get:

$\small{1~=~(Ax+B) (x-1)~+~A_1 (x^2+1)}$

7. After equating the numerators, we can use suitable substitution.

   ♦ Put x = 1. We get: $\small{1~=~(Ax+B) (x-1)~+~A_1 (x^2+1)}$. So A₁ = 1/2
   
   ♦ Put x = 0. We get: 1 = −B + 1/2. So B = −1/2
   
   ♦ Put x = −1. We get: 1 = (−A − (1/2))(-2) + (1/2)2. So A = −(1/2)

8. Now the result in (4) becomes:

$\small{\frac{1}{(x^2 + 1)(x - 1)}~=~\left[\frac{(-1/2)x ~-~ (1/2)}{x^2 + 1}\right]~+~\frac{1/2}{ x - 1}}$

$\small{~=~\left[\frac{-x ~-~ 1}{2(x^2 + 1)}\right]~+~\frac{1}{2(x - 1)}}$

$\small{~=~\frac{1}{2(x - 1)}~-~\frac{x}{2(x^2 + 1)}~-~\frac{1}{2(x^2 + 1)}}$

9. So based on the result in (2), we get:

$\small{\frac{x^4}{(x-1)(x^2+1)}\,=\,(x+1)~+~ \frac{1}{2(x - 1)}~-~\frac{x}{2(x^2 + 1)}~-~\frac{1}{2(x^2 + 1)}}$

10. Now the integration becomes easy. We get:

$\small{\int \left[\frac{x^4}{(x-1)(x^2+1)} \right]dx}$

$\small{~=~\frac{x^2}{2}~+~x~+~\frac{1}{2}\,\log \left|x-1 \right|~-~\frac{1}{4} \log (x^2+1)~-~\frac{1}{2} \tan^{-1}x~+~\rm{C}}$

• The reader may write all the steps involved in the integration process.

Solved Example 23.122
Find $\small{\int \left[\log(\log x)+\frac{1}{(\log x)^2} \right]dx}$
Solution:
1. Let $\small{I=\int \left[\log(\log x)+\frac{1}{(\log x)^2} \right]dx}$

$\small{~=~\int \left[\log(\log x) \right]dx~+~\int \left[\frac{1}{(\log x)^2} \right]dx~=~I_1 + I_2}$

2. First we will calculate I₁:

(a) Assigning first and second functions:

   ♦ Let first function be: $\small{f(x)=\log(\log x)}$

   ♦ Let second function be: $\small{g(x)=1}$

(b) Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[1\right]dx}~=~x}$

(c) $\small{\big[f(x) \left(A \right) \big]~=~\big[\log (\log x) \, \left(x \right) \big]}$

• This is the first term.

(d) $\small{f'(x)~=~\frac{1}{\log (x)}\left(\frac{1}{x} \right)~=~\frac{1}{x \log(x)}}$

(e) $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{1}{x \log(x)}\,\left(x \right)\big]dx}~=~\int{\big[\frac{1}{\log(x)}\big]dx}}$

• This is the second term.

(f) So we get:

$\small{\int{\left[\log(\log x) \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[\left(x \right) \log (\log x)\big]~-~\int{\big[\frac{1}{\log(x)}\big]dx}}$

3. In the above result, let us calculate $\small{\int{\big[\frac{1}{\log(x)}\big]dx}}$

(a) Assigning first and second functions:

   ♦ Let first function be: $\small{f(x)=\frac{1}{\log(x)}}$

   ♦ Let second function be: $\small{g(x)=1}$

(b) Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[1\right]dx}~=~x}$

(c) $\small{\big[f(x) \left(A \right) \big]~=~\big[\frac{1}{\log(x)} \, \left(x \right) \big]}$

• This is the first term.

(d) $\small{f'(x)~=~\frac{d}{dx}\left([\log(x)]^{-1} \right)~=~(-1)[\log(x)]^{-2}\left(\frac{1}{x} \right)~=~\frac{-1}{x[\log(x)]^2}}$

(e) $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{-1}{x[\log(x)]^2}\,\left(x \right)\big]dx}~=~\int{\big[\frac{-1}{[\log(x)]^2}\big]dx}}$

• This is the second term.

(f) So we get:

$\small{\int{\left[\frac{1}{\log(x)} \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[\frac{x}{\log(x)}\big]~-~\int{\big[\frac{-1}{[\log(x)]^2}\big]dx}}$

$\small{~=~\big[\frac{x}{\log(x)}\big]~+~\int{\big[\frac{1}{[\log(x)]^2}\big]dx}}$

4. Substituting the above result in (2), we get:

$\small{\int{\left[\log(\log x) \right]dx}~=~\big[\left(x \right) \log (\log x)\big]~-~\int{\big[\frac{1}{\log(x)}\big]dx}}$

$\small{\Rightarrow\int{\left[\log(\log x) \right]dx}~=~\big[\left(x \right) \log (\log x)\big]~-~\Bigg[\big[\frac{x}{\log(x)}\big]~+~\int{\big[\frac{1}{[\log(x)]^2}\big]dx} \Bigg]}$

$\small{\Rightarrow I_1 =\int{\left[\log(\log x) \right]dx}~=~\big[\left(x \right) \log (\log x)\big]~-~\big[\frac{x}{\log(x)}\big]~-~\int{\big[\frac{1}{(\log x)^2}\big]dx}}$

5. Now we can substitute the value of I₁ in step (1). We get:

$\small{I=I_1 + I_2}$

$\small{=\int \left[\log(\log x) \right]dx~+~\int \left[\frac{1}{(\log x)^2} \right]dx}$

$\small{=\big[\left(x \right) \log (\log x)\big]~-~\big[\frac{x}{\log(x)}\big]~-~\int{\big[\frac{1}{(\log x)^2}\big]dx}~+~\int \left[\frac{1}{(\log x)^2} \right]dx}$

$\small{=\big[\left(x \right) \log (\log x)\big]~-~\big[\frac{x}{\log(x)}\big]~+~\rm{C}}$

Solved Example 23.123
Find $\small{\int{\left[\frac{\sin(2x) \cos(2x)}{\sqrt{9 - \cos^4(2x)}} \right]dx}}$
Solution:
1.1. The derivative of $\small{\cos^2(2x)}$ is $\small{-4 \sin(2x) \cos(2x)}$.

So we put $\small{u = \cos^2(2x)}$

$\small{\Rightarrow \frac{du}{dx} = -4 \sin(2x) \cos(2x)}$
$\small{\Rightarrow -4 \sin(2x) \cos(2x) dx = du}$

2. So we want:
$\small{\int \left[\frac{(-4)\sin(2x) \cos(2x)}{(-4)\sqrt{9 - \cos^4(2x)}} \right]dx~=~\int \left[\frac{1}{(-4)\sqrt{9 - u^2}} \right]du}$

$\small{~=~\left(\frac{-1}{4} \right)\int \left[\frac{1}{\sqrt{9 - u^2}} \right]du~=~\left(\frac{-1}{4} \right)\int \left[\frac{1}{\sqrt{3^2 - u^2}} \right]du}$

3. This is a standard integral. We can write:

$\small{\left(\frac{-1}{4} \right)\int \left[\frac{1}{\sqrt{3^2 - u^2}} \right]du~=~\frac{-1}{4} \left[\sin^{-1}\frac{u}{3} \right]+\rm{C}}$

4. Substituting for u, we get:
$\small{\frac{-1}{4} \left[\sin^{-1}\left(\frac{\cos^2(2x)}{3} \right) \right]+\rm{C}}$

Solved Example 23.124
Evaluate $\small{\int_{-1}^{\frac{3}{2}}{\left[\left|x \sin(\pi x) \right| \right]dx}}$
Solution:
1. Let us determine the intervals where $\small{x \sin(\pi x)}$ is +ve or -ve.

The given interval is [−1,1.5]. This interval can be split into three: [−1,0], [0,1] and [1,1.5]

2. Consider the interval [−1,0]. In this interval, all x values are −ve. So $\small{(\pi x)}$ is −ve. It corresponds to III and IV quadrants, where sine is −ve. So $\small{x \sin(\pi x)}$ is +ve.

So for this interval, we can write:

$\small{\left|x \sin(\pi x) \right|=x \sin(\pi x)~\text{if}~-1 < x < 0}$

3. Consider the interval [0,1]. In this interval, all x values are +ve. So $\small{(\pi x)}$ is +ve. It corresponds to I and II quadrants, where sine is +ve. So $\small{x \sin(\pi x)}$ is +ve.

So for this interval, we can write:

$\small{\left|x \sin(\pi x) \right|=x \sin(\pi x)~\text{if}~0 < x < 1}$

• (2) and (3) give the same result. Also, they are adjacent intervals. So those two intervals can be combined.

4. Consider the interval [1,1.5]. In this interval, all x values are +ve. So $\small{(\pi x)}$ is +ve. It corresponds to III quadrant, where sine is −ve. So $\small{x \sin(\pi x)}$ is −ve.

So for this interval, we can write:

$\small{\left|x \sin(\pi x) \right|=-x \sin(\pi x)~\text{if}~1 < x < 1.5}$

5. Consider the exact points x = −1, x = 0, x = 1 and x = 1.5

(a) When x = −1, $\small{x \sin(\pi x)} = (-1)\sin(-\pi)=0$

So for this point, we can write:

$\small{\left|x \sin(\pi x) \right|= \pm x \sin(\pi x)~\text{if}~ x = -1}$

(b) When x = 0, $\small{x \sin(\pi x)} = (0)\sin(0)=0$

So for this point, we can write:

$\small{\left|x \sin(\pi x) \right|= \pm x \sin(\pi x)~\text{if}~ x = 0}$

(c) When x = 1, $\small{x \sin(\pi x)} = (1)\sin(\pi)=0$

So for this point, we can write:

$\small{\left|x \sin(\pi x) \right|= \pm x \sin(\pi x)~\text{if}~ x = 1}$

(d) When x = 1.5, $\small{x \sin(\pi x)} = (1.5)\sin((1.5)\pi)=-1.5$

So for this point, we can write:

$\small{\left|x \sin(\pi x) \right|= -x \sin(\pi x)~\text{if}~ x = 1.5}$

6. Combining (2), (3), (4) and (5), we can write:

$\left|x \sin(\pi x) \right| = \begin{cases} x \sin(\pi x),  & \text{if}~-1 \le x \le 1  \\[1.5ex] -x \sin(\pi x), & \text{if}~~1 \le x \le 1.5  \end{cases}$

7. The integrand changes at 1. So we will split the interval at that point. Then by applying P₂, it can be written as:

$\small{\int_{-1}^{1.5}{\left[\left|x \sin(\pi x) \right| \right]dx}}$

$\small{\,=\,\int_{-1}^{1}{\left[\left|x \sin(\pi x) \right| \right]dx}\,+\,\int_{1}^{1.5}{\left[\left|x \sin(\pi x) \right| \right]dx}}$

• We will denote it as:

$\small{\int_{-1}^{1.5}{\left[\left|x \sin(\pi x) \right| \right]dx}\,=\,I_1\,+\,I_2}$

8. Choosing the appropriate segments:

• For I₁, we must use the segment: $\small{\left|x \sin(\pi x) \right|\,=\,x \sin(\pi x),~\text{if}~-1 \le x \le 1}$

• For I₂, we must use the segment: $\small{\left|x \sin(\pi x) \right|\,=\,-x \sin(\pi x),~\text{if}~1 \le x \le 1.5}$

9. So from (7), we get:

$\small{\int_{-1}^{1.5}{\left[\left|x \sin(\pi x) \right| \right]dx}=I_1 + I_2}$

$\small{\,=\,\int_{-1}^{1}{\left[x \sin(\pi x) \right]dx}\,+\,\int_{1}^{1.5}{\left[x \sin(\pi x) \right]dx}}$

$\small{\,=\,\int_{-1}^{1}{\left[x \sin(\pi x) \right]dx}\,-\,\int_{1}^{1.5}{\left[x \sin(\pi x) \right]dx}~=~I_3 ~-~ I_4}$

10. So our next task is to find the indefinite integral: $\small{\int{\left[x \sin(\pi x) \right]dx}}$   

(a) Assigning first and second functions:

   ♦ Let first function be: $\small{f(x)= x}$

   ♦ Let second function be: $\small{g(x)=\sin(\pi x)}$

(b) Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[\sin(\pi x)\right]dx}~=~\frac{-\cos(\pi x)}{\pi}}$

(c) $\small{\big[f(x) \left(A \right) \big]~=~\big[x \, \left(\frac{-\cos(\pi x)}{\pi} \right) \big]~=~\big[\frac{-x \cos(\pi x)}{\pi}  \big]}$

• This is the first term.

(d) $\small{f'(x)~=~1}$

(e) $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[(1)\,\left(\frac{-\cos(\pi x)}{\pi} \right)\big]dx}~=~\frac{-1}{\pi}\int{\big[\cos (\pi x)\big]dx}}$

$\small{~=~\frac{-1}{\pi} \left(\frac{\sin(\pi x)}{\pi} \right)~=~\frac{-\sin(\pi x)}{\pi^2}}$

• This is the second term.

(f) So we get:

$\small{\int{\left[x \sin(\pi x) \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[\frac{-x \cos(\pi x)}{\pi}\big]~+~\big[\frac{\sin(\pi x)}{\pi^2}\big]}$

11. Now we can calculate I₃

$\small{\int_{-1}^{1}{\left[x \sin(\pi x) \right]dx}~=~\big[\frac{-x \cos(\pi x)}{\pi}~+~\frac{\sin(\pi x)}{\pi^2}\big]_{-1}^1}$

$\small{~=~\big[\left(\frac{-(1) \cos(\pi)}{\pi}~+~\frac{\sin(\pi)}{\pi^2} \right)~-~\left(\frac{-(-1) \cos(-\pi)}{\pi}~+~\frac{\sin(-\pi)}{\pi^2} \right)\big]}$

$\small{~=~\big[\left(\frac{-(1) (-1)}{\pi}~+~\frac{0}{\pi^2} \right)~-~\left(\frac{-(-1) (-1)}{\pi}~+~\frac{0}{\pi^2} \right)\big]}$

$\small{~=~\big[\left(\frac{1}{\pi} \right)~-~\left(\frac{(-1)}{\pi} \right)\big]~=~\frac{2}{\pi}}$

12. Next we can calculate I₄

$\small{\int_{1}^{1.5}{\left[x \sin(\pi x) \right]dx}~=~\big[\frac{-x \cos(\pi x)}{\pi}~+~\frac{\sin(\pi x)}{\pi^2}\big]_{1}^{1.5}}$

$\small{~=~\big[\left(\frac{-(1.5) \cos(1.5\pi)}{\pi}~+~\frac{\sin(1.5\pi)}{\pi^2} \right)~-~\left(\frac{-(1) \cos(\pi)}{\pi}~+~\frac{\sin(\pi)}{\pi^2} \right)\big]}$

$\small{~=~\big[\left(\frac{0}{\pi}~+~\frac{-1}{\pi^2} \right)~-~\left(\frac{-(1) (-1)}{\pi}~+~\frac{0}{\pi^2} \right)\big]}$

$\small{~=~\big[\left(\frac{-1}{\pi^2} \right)~-~\left(\frac{1}{\pi} \right)\big]~=~\frac{-1}{\pi^2}~-~\frac{1}{\pi}}$

13. Thus we get: I₃ − I₄

$\small{~=~\frac{2}{\pi}~-~\left(\frac{-1}{\pi^2}~-~\frac{1}{\pi} \right)}$

$\small{~=~\frac{2}{\pi}~+~\frac{1}{\pi^2}~+~\frac{1}{\pi}}$

$\small{~=~\frac{3}{\pi}~+~\frac{1}{\pi^2}}$

Solved Example 23.125
Evaluate $\small{\int_{1}^{4}{\left[\left|x-1 \right| ~+~\left|x-2 \right|~+~\left|x-3 \right| \right]dx}}$
Solution:
1. We can write:

$\small{\int_{1}^{4}{\left[\left|x-1 \right| ~+~\left|x-2 \right|~+~\left|x-3 \right| \right]dx}}$

= $\small{\int_{1}^{4}{\left[\left|x-1 \right|  \right]dx}~+~\int_{1}^{4}{\left[\left|x-2 \right| \right]dx}~+~\int_{1}^{4}{\left[\left|x-3 \right| \right]dx}}$

= I₁ + I₂ + I₃

2. First we will evaluate I₁:

(a) Let us determine the intervals where (x−1) is +ve or −ve. For that, first we solve the inequality: x−1<0

We have: $\small{x-1 < 0}$

$\small{\Rightarrow x < 1}$

• So when x is less than 1, (x−1) will be −ve.

• That means, when x is less than 1, $\small{\left|x-1 \right|\,=\,-(x-1)}$

(b) Next we solve the inequality:x−1>0

We have: $\small{x-1 > 0}$

$\small{\Rightarrow x > 1}$

• So when x is greater than 1, (x-1) will be +ve.

• That means, when x is greater  than 1, $\small{\left|x-1 \right|\,=\,x-1}$

(c) Also, we must solve the equation x-1 = 0

This gives x = 1

• So when x is equal to 1, (x-1) will be zero.

• That means, when x is equal to 1, $\small{\left|x-1 \right|\,=\,\pm(x-1)}$

(d) Now we can write a piece wise function:

$\left|x-1 \right| = \begin{cases} x-1,  & \text{if}~x \ge 1 \\[1.5ex] -(x-1), & \text{if}~x<1  \end{cases}$

(e) The given interval is [1,4]. So x is always greater than or equal to 1. Therefore, we need to consider the first segment only.

We can write:

$\small{I_1~=~\int_{1}^{4}{\left[\left|x-1 \right| \right]dx}\,=\,\int_{1}^{4}{\left[x-1 \right]dx}}$

$\small{\,=\,\left[\frac{x^2}{2}\,-\,x \right]_{1}^{4}}$

$\small{\,=\,\left[\frac{16}{2}\,-\,4\,-\,\left(\frac{1}{2}\,-\,1 \right) \right]}$

$\small{\,=\,\left[4\,-\,\left(\frac{-1}{2} \right) \right]~=~\frac{9}{2}}$

3. Next we will evaluate I₂:

(a) Let us determine the intervals where (x−2) is +ve or −ve. For that, first we solve the inequality: x−2<0

We have: $\small{x-2 < 0}$

$\small{\Rightarrow x < 2}$

• So when x is less than 1, (x−2) will be −ve.

• That means, when x is less than 2, $\small{\left|x-2 \right|\,=\,-(x-1)}$

(b) Next we solve the inequality:x−2>0

We have: $\small{x-2 > 0}$

$\small{\Rightarrow x > 2}$

• So when x is greater than 2, (x-2) will be +ve.

• That means, when x is greater  than 2, $\small{\left|x-2 \right|\,=\,x-2}$

(c) Also, we must solve the equation x-2 = 0

This gives x = 2

• So when x is equal to 2, (x-2) will be zero.

• That means, when x is equal to 2, $\small{\left|x-2 \right|\,=\,\pm(x-2)}$

(d) Now we can write a piece wise function:

$\left|x-2 \right| = \begin{cases} x-2,  & \text{if}~x \ge 2 \\[1.5ex] -(x-2), & \text{if}~x<2  \end{cases}$

(e) The integrand changes at 2. So we will split the interval at 2. Then by applying P₂, it can be written as:

$\small{\int_{1}^{4}{\left[\left|x-2 \right| \right]dx}\,=\,\int_{1}^{2}{\left[\left|x-2 \right| \right]dx}\,+\,\int_{2}^{4}{\left[\left|x-2 \right| \right]dx}}$

• We will denote it as:

$\small{\int_{1}^{4}{\left[\left|x-2 \right| \right]dx}\,=\,I_4\,+\,I_5}$

(f) Choosing the appropriate segments:

• For I₄, we must use the segment: $\small{\left|x-2 \right|\,=\,-(x-2),~\text{if}~x<2}$

• For I₅, we must use the segment: $\small{\left|x-2 \right|\,=\,x-2,~\text{if}~x \ge 2}$

(g) So from (e), we get:

$\small{\int_{1}^{4}{\left[\left|x-2 \right| \right]dx}\,=\,\int_{1}^{2}{\left[-(x-2) \right]dx}\,+\,\int_{2}^{4}{\left[x-2 \right]dx}}$

$\small{\,=\,(-1)\int_{1}^{2}{\left[x-2 \right]dx}\,+\,\int_{2}^{4}{\left[x-2 \right]dx}}$

$\small{\,=\,(-1)\left[\frac{x^2}{2}\,-\,2x \right]_{1}^{2}\,+\,\left[\frac{x^2}{2}\,-\,2x \right]_{2}^{4}}$

$\small{\,=\,(-1)\left[\frac{2^2}{2}\,-\,4\,-\,\left(\frac{1^2}{2}\,-\,2 \right) \right]\,+\,\left[\frac{4^2}{2}\,-\,8\,-\,\left(\frac{2^2}{2}\,-\,4 \right) \right]}$

$\small{\,=\,(-1)\left[-2\,-\,\left(\frac{-3}{2} \right) \right]\,+\,\left[0\,-\,\left(-2 \right) \right]}$

$\small{\,=\,(-1)\left[\frac{-1}{2}  \right]\,+\,\left[2  \right]~=~\frac{5}{2}}$

4. In a similar way, we can calculate I₃. We get:

$\small{\int_{1}^{4}{\left[\left|x-3 \right| \right]dx}~=~\frac{5}{2}}$ 

5. Therefore, we can write:

$\small{I = I_1 + I_2 + I_3 ~=~\frac{9}{2} + \frac{5}{2} + \frac{5}{2}~=~\frac{19}{2}}$

Solved Example 23.126
Evaluate $\small{\int_{0}^{\pi}{\left[\frac{x}{a^2 \cos^2 x~+~b^2 \sin^2 x} \right]dx}}$
Solution:
1. Applying P₄, we can write:

$\small{I=\int_{0}^{\pi}{\left[\frac{x}{a^2 \cos^2 x~+~b^2 \sin^2 x} \right]dx}=\int_{0}^{\pi}{\left[\frac{\pi - x}{a^2 \cos^2 (\pi - x)~+~b^2 \sin^2 (\pi - x)} \right]dx}}$

$\small{=\int_{0}^{\pi}{\left[\frac{\pi-x}{a^2 \cos^2 x~+~b^2 \sin^2 x} \right]dx}}$

$\small{=\int_{0}^{\pi}{\left[\frac{\pi}{a^2 \cos^2 x~+~b^2 \sin^2 x} \right]dx}~-~\int_{0}^{\pi}{\left[\frac{x}{a^2 \cos^2 x~+~b^2 \sin^2 x} \right]dx}}$

$\small{=\int_{0}^{\pi}{\left[\frac{\pi}{a^2 \cos^2 x~+~b^2 \sin^2 x} \right]dx}~-~I}$

2. So we can write: $\small{2I=\int_{0}^{\pi}{\left[\frac{\pi}{a^2 \cos^2 x~+~b^2 \sin^2 x} \right]dx}}$

$\small{\Rightarrow I=\frac{\pi}{2} \int_{0}^{\pi}{\left[\frac{1}{a^2 \cos^2 x~+~b^2 \sin^2 x} \right]dx}}$

3. Here we can apply P₆. The reader may write how P₆ is applicable.

We get: $\small{I=\frac{\pi}{2} (2) \int_{0}^{\frac{\pi}{2}}{\left[\frac{1}{a^2 \cos^2 x~+~b^2 \sin^2 x} \right]dx}=\pi \int_{0}^{\frac{\pi}{2}}{\left[\frac{1}{a^2 \cos^2 x~+~b^2 \sin^2 x} \right]dx}}$

4. Dividing both numerator and denominator by $\small{\cos^2 x}$, we get:

$\small{I=\pi \int_{0}^{\frac{\pi}{2}}{\left[\frac{\sec^2 x}{a^2~+~b^2 \tan^2 x} \right]dx}}$

• Put $\small{u=b \tan x \Rightarrow \frac{du}{dx} = b \sec^2 x \Rightarrow b \sec^2 x dx = du}$

   ♦ Also, when x approach zero, u approach zero

   ♦ And when x approach $\small{\frac{\pi}{2}}$, u approach $\small{\infty}$

• So we want to evaluate:

$\small{I= \frac{\pi}{b} \int_{0}^{\frac{\pi}{2}}{\left[\frac{b \sec^2 x}{a^2~+~b^2 \tan^2 x} \right]dx} = \frac{\pi}{b} \int_{0}^{\infty}{\left[\frac{1}{a^2~+~u^2} \right]du}}$

5. This is a standard integral. We get:

$\small{I= \frac{\pi}{b} \left[\frac{1}{a} \tan^{-1} \frac{u}{a} \right]_0^{\infty} = \frac{\pi}{ab} \left[ \frac{\pi}{2}~-~0 \right]~=~\frac{\pi^2}{2ab}}$

Solved Example 23.127
Integrate $\small{\frac{1}{x - x^3}}$
Solution:
1. First we will rearrange the given expression:
$\small{\frac{1}{x - x^3}~=~\frac{1}{(x)\frac{x^3}{x^3} ~-~ (x^3)\frac{x^3}{x^3}}~=~\frac{1}{x^3 \left[(x)\frac{1}{x^3} ~-~ (x^3)\frac{1}{x^3} \right]}~=~\frac{1}{x^3 \left[\frac{1}{x^2} ~-~ 1 \right]}}$

2. The derivative of $\small{\left(\frac{1}{x^2} - 1 \right)}$ is $\small{\frac{-2}{x^3}}$

So we put $\small{u=\frac{1}{x^2} - 1}$

$\small{\Rightarrow \frac{du}{dx}~=~\frac{-2}{x^3}~~~\Rightarrow \frac{-2}{x^3}dx~=~du}$

3. So we want: $\small{I = \int{\left[\frac{1}{x - x^3} \right]dx}}$

$\small{=\int{\left[\frac{1}{x^3 \left[\frac{1}{x^2} ~-~ 1 \right]} \right]dx}=\int{\left[\frac{(-2)}{(-2)x^3 \left[\frac{1}{x^2} ~-~ 1 \right]} \right]dx}=\int{\left[\frac{1}{(-2) \left[u \right]} \right]du}}$

$\small{= \frac{-1}{2} \int{\left[\frac{1}{u} \right]du}~=~\frac{-\log u}{2}+\rm{C}~=~\frac{1}{2} \log \left(\frac{1}{u} \right)+\rm{C}}$

4. We wrote: $\small{u=\frac{1}{x^2} - 1~=~\frac{1 - x^2}{x^2}}$

• So $\small{\frac{1}{u}~=~\frac{x^2}{1 - x^2}}$

• Substituting this in (3), we get:

$\small{I = \int{\left[\frac{1}{x - x^3} \right]dx}~=~\frac{1}{2} \log \left|\frac{x^2}{1 - x^2} \right| +\rm{C}}$

Solved Example 23.128
Integrate $\small{\frac{1}{x \sqrt{ax - x^2}}}$
Solution:
1. First we will rearrange the given expression:
$\small{\frac{1}{x \sqrt{ax - x^2}}
~=~\frac{\left(ax - x^2 \right)^{-(1/2)}}{x}
~=~\frac{\left[ax \left(\frac{x^2}{x^2} \right)- x^2\left(\frac{x^2}{x^2} \right) \right]^{-(1/2)}}{x}
}$

$\small{
~=~\frac{(x^2)^{-(1/2)} \left[ax \left(\frac{1}{x^2} \right)- x^2\left(\frac{1}{x^2} \right) \right]^{-(1/2)}}{x}
~=~\frac{x^{-1} \left[\frac{a}{x}-1   \right]^{-(1/2)}}{x}
}$

$\small{
~=~\frac{\left[\frac{a}{x}-1  \right]^{-(1/2)}}{x^2}
}$

2. Put $\small{u = \frac{a}{x}-1}$. Then we get:

$\small{\frac{du}{dx}=\frac{-a}{x^2}}$

$\small{\Rightarrow \frac{-a}{x^2} dx~=~du}$

3. So we want: $\small{I = \int{\left[\frac{\left[\frac{a}{x}-1  \right]^{-(1/2)}}{x^2} \right]dx}= \int{\left[\frac{(-a)\left[\frac{a}{x}-1  \right]^{-(1/2)}}{(-a)x^2} \right]dx}}$

$\small{=\int{\left[\frac{\left[u  \right]^{-(1/2)}}{(-a)} \right]du}=\left(\frac{-1}{a} \right)\int{\left[u^{-(1/2)} \right]du}}$

4. This integration gives:

$\small{\left(\frac{-1}{a} \right)\left[\frac{u^{1/2}}{1/2} ~+~\rm{C_1}\right]=\left[\frac{2 u^{1/2}}{(-a)} ~+~\rm{C}\right] =\frac{-2 u^{1/2}}{a}~+~\rm{C}}$

5. Substituting for u, we get:

$\small{I= \frac{-2 }{a} \left(\frac{a}{x}-1 \right)^{1/2}~+~\rm{C}}$

Solved Example 23.129
Integrate $\small{\frac{1}{x^2(x^4 + 1)^{3/4}}}$
Solution:
1. First we will rearrange the given expression:
$\small{\frac{1}{x^2(x^4 + 1)^{3/4}}
~=~\frac{\left(x^4 + 1 \right)^{-(3/4)}}{x^2}
~=~\frac{\left[x^4 \left(\frac{x^4}{x^4} \right)+ 1\left(\frac{x^4}{x^4} \right) \right]^{-(3/4)}}{x^2}
}$

$\small{
~=~\frac{(x^4)^{-(3/4)} \left[x^4 \left(\frac{1}{x^4} \right)+ 1\left(\frac{1}{x^4} \right) \right]^{-(3/4)}}{x^2}
~=~\frac{x^{-3} \left[1+ \frac{1}{x^4}  \right]^{-(3/4)}}{x^2}
}$

$\small{
~=~\frac{\left[1+ \frac{1}{x^4}  \right]^{-(3/4)}}{x^5}
}$

2. Put $\small{u = 1+ \frac{1}{x^4}}$. Then we get:

$\small{\frac{du}{dx}=\frac{-4}{x^5}}$

$\small{\Rightarrow \frac{-4}{x^5} dx~=~du}$

3. So we want: $\small{I = \int{\left[\frac{\left[1+ \frac{1}{x^4}  \right]^{-(3/4)}}{x^5} \right]dx}= \int{\left[\frac{(-4)\left[1+ \frac{1}{x^4}  \right]^{-(3/4)}}{(-4)x^5} \right]dx}}$

$\small{=\int{\left[\frac{\left[u  \right]^{-(3/4)}}{(-4)} \right]du}=\left(\frac{-1}{4} \right)\int{\left[u^{-(3/4)} \right]du}}$

4. This integration gives:

$\small{\left(\frac{-1}{4} \right)\left[\frac{u^{1/4}}{1/4} ~+~\rm{C_1}\right]=\left[\frac{u^{1/4}}{(-1)} ~+~\rm{C}\right] =(-1)u^{1/4}~+~\rm{C}}$

5. Substituting for u, we get:

$\small{I= (-1)\left(1+ \frac{1}{x^4} \right)^{1/4}~+~\rm{C}}$

Solved Example 23.130
Integrate $\small{\frac{1}{x^{\frac{1}{2}} + x^{\frac{1}{3}}}}$
Solution:
1. First we will rearrange the given expression:
$\small{\frac{1}{x^{1/2} + x^{1/3}}
~=~\frac{1}{(x^{1/2})\frac{x^{1/3}}{x^{1/3}} ~+~ (x^{1/3})\frac{x^{1/3}}{x^{1/3}}}
}$

$\small{
~=~\frac{1}{x^{1/3} \left[(x^{1/2})\frac{1}{x^{1/3}} ~+~ (x^{1/3})\frac{1}{x^{1/3}} \right]}
~=~\frac{1}{x^{1/3} \left[x^{1/6} ~+~ 1 \right]}}$

2. Put $\small{u = x^{1/6}}$. Then we get:

$\small{\frac{du}{dx}=\frac{1}{6} x^{-5/6}=\frac{1}{6\, x^{5/6}}}$

$\small{\Rightarrow dx = 6\, x^{5/6} du}$

• Also, $\small{x = u^6}$, $\small{x^{1/3} = u^2}$ and $\small{x^{5/6} = u^5}$

3. So we want: $\small{I = \int{\left[\frac{1}{x^{1/2} + x^{1/3}} \right]dx}}$

$\small{=\int{\left[\frac{1}{x^{1/3} \left[x^{1/6} ~+~ 1 \right]} \right]dx}=\int{\left[\frac{6\, x^{5/6}}{u^{2} \left[u ~+~ 1 \right]} \right]du}}$

$\small{=\int{\left[\frac{6\, u^{5}}{u^{2} \left[u ~+~ 1 \right]} \right]du}=\int{\left[\frac{6\, u^{3}}{ u ~+~ 1 } \right]du}=6 \int{\left[\frac{u^{3}}{ u ~+~ 1 } \right]du}}$

4. So our next task is to integrate: $\small{\frac{u^{3}}{ u ~+~ 1 }}$

(a) The numerator is a polynomial of degree 3. The denominator is a polynomial of degree 1.

(b) So it is not a proper rational function. We must do long division. We get:

$\small{\frac{u^3}{u+1}\,=\,(u^2 - u +1)~+~ \frac{1}{(u + 1)}}$

• The reader may write all steps involved in the long division (or any other suitable method) process.

5. So we can write:

$\small{I~=~6 \int{\left[\frac{u^{3}}{ u ~+~ 1 } \right]du}=6 \int{\left[u^2 - u +1~+~ \frac{1}{u + 1} \right]du}}$

$\small{=6 \left[\frac{u^3}{3} - \frac{u^2}{2} + u ~+~ \log \left|u+1 \right| ~+~\rm{C_1}\right]}$

$\small{= 2 u^3 - 3 u^2 + 6u ~+~ 6\log \left|u+1 \right| ~+~\rm{6C_1}}$

6. Substituting for u, we get:

$\small{I= 2 \left(x^{1/6} \right)^3 - 3 \left(x^{1/6} \right)^2 + 6\left(x^{1/6} \right) ~+~ 6\log \left|\left(x^{1/6} \right)+1 \right| ~+~\rm{6C_1}}$

$\small{= 2 x^{1/2}  - 3 x^{1/3}  + 6x^{1/6}  ~+~ 6\log \left|x^{1/6}+1 \right| ~+~\rm{C}}$

Solved Example 23.131
Integrate $\small{\frac{x^3}{\sqrt{1 - x^8}}}$
Solution:
1. First we will rearrange the given expression:
$\small{\frac{x^3}{\sqrt{1 - x^8}}
~=~\frac{x^3}{\sqrt{1 - (x^4)^2}}}$

2. Put $\small{u = x^4}$. Then we get:

$\small{\frac{du}{dx}=4x^3}$

$\small{\Rightarrow 4x^3 dx~=~du}$

3. So we want: $\small{I = \int{\left[\frac{x^3}{\sqrt{1 - (x^4)^2}}\right]dx}= \int{\left[\frac{4 x^3}{4 \sqrt{1 - (x^4)^2}} \right]dx}}$

$\small{=\int{\left[\frac{1}{4 \sqrt{1-u^2}} \right]du}=\left(\frac{1}{4} \right)\int{\left[\frac{1}{\sqrt{1-u^2}} \right]du}}$

4. This is a standard integration. It gives:

$\small{\left(\frac{1}{4} \right)\left[\sin^{-1}u ~+~\rm{C_1}\right]=\frac{\sin^{-1}u}{4} ~+~\rm{C}}$

5. Substituting for u, we get:

$\small{I= \frac{\sin^{-1}(x^4)}{4}~+~\rm{C}}$

Solved Example 23.132
Integrate $\small{\frac{\cos x}{\sqrt{4 - \sin^2(x)}}}$
Solution:
1. First we will rearrange the given expression:
$\small{\frac{\cos x}{\sqrt{4 - \sin^2(x)}}
~=~\frac{\cos x}{\sqrt{2^2 - \sin^2(x)}}}$

2. Put $\small{u = \sin x}$. Then we get:

$\small{\frac{du}{dx}=\cos x}$

$\small{\Rightarrow \cos x\, dx~=~du}$

3. So we want: $\small{I = \int{\left[\frac{\cos x}{\sqrt{2^2 - \sin^2(x)}}\right]dx}= \int{\left[\frac{1}{\sqrt{2^2 - u^2}} \right]du}}$

4. This is a standard integration. It gives:

$\small{\sin^{-1}\frac{u}{2} ~+~\rm{C}}$

5. Substituting for u, we get:

$\small{I= \sin^{-1} \left(\frac{\sin x}{2} \right)~+~\rm{C}}$

---

In the next section, we will see a few more miscellaneous examples.

Previous

Contents

Next 

Copyright©2025 Higher secondary mathematics.blogspot.com