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.

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23.28 - More Solved Examples on Properties of Definite Integrals

In the previous section, we saw some properties of definite integrals. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved Example 23.112
Evaluate $\small{\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \right]dx}}$
Solution:
1. The limits are $\small{0~\rm{and}~\frac{\pi}{2}}$. So we will apply:

$\bf{{\rm{P_4}}:\int_0^a{\left[f(x) \right]dx}~=~\int_0^a{\left[f(a-x) \right]dx}}$

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

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

2. Now, I + I = 2I =

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

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

Therefore, $\small{I = \frac{\pi}{4}}$

Solved Example 23.113
Evaluate $\small{\int_{0}^{a}{\left[\frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}} \right]dx}}$
Solution:
1. The limits are $\small{0~\rm{and}~a}$. So we will apply:

$\bf{{\rm{P_4}}:\int_0^a{\left[f(x) \right]dx}~=~\int_0^a{\left[f(a-x) \right]dx}}$

We get:
$\small{I~=~\int_{0}^{a}{\left[\frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}}\right]dx}~=~\int_{0}^{a}{\left[\frac{\sqrt{a-x}}{\sqrt{a-x} ~+~ \sqrt{a-(a-x)}}\right]dx}  }$

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

2. Now, I + I = 2I =

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

$\small{~=\int_{0}^{a}{\left[1 \right]dx}=[x]_{0}^{a}=a}$

Therefore, $\small{I = \frac{a}{2}}$

Solved Example 23.114
Evaluate $\small{\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}{\left[\sin^7 x \right]dx}}$
Solution:
1. The limits are $\small{\frac{-\pi}{2}~\rm{and}~\frac{\pi}{2}}$.
So we check whether $\small{f(x)=\sin^7 x}$ is even or odd.

(a) $\small{f(-x)\,=\,\sin^7 (-x)\,=\,\left(\sin(-x) \right)^7}$

$\small{\,=\,\left(-\sin(x) \right)^7\,=\,-\sin^7 x}$

$\small{\because \sin(-x) = -\sin x}$

(b) $\small{-f(x)\,=\,-\sin^7 (x)}$

(c) We see that:

$\small{f(-x) \,=\,-f(x)}$

• So the given f(x) is an odd function.

2. Since the given function is odd, we can apply:

$\bf{{\rm{P_7~(ii)}}:}$

$\bf{\int_{-a}^{a}{\left[f(x) \right]dx}~=~0}$
   ♦ If $\small{f}$ is an odd function

• We can write: $\small{\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}{\left[\sin^7 x \right]dx}~=~0}$

Solved Example 23.115
Evaluate $\small{\int_{0}^{2\pi}{\left[\cos^5 x \right]dx}}$
Solution:
1. The limits are 0 and $\small{2 \pi}$.
So we can apply:

$\bf{{\rm{P_6}}:}$

$\bf{\int_0^{2a}{\left[f(x) \right]dx}~=~2 \int_0^a{\left[f(x) \right]dx}}$
   ♦ If $\small{f(2a - x)~=~f(x)}$

$\bf{\int_0^{2a}{\left[f(x) \right]dx}~=~0}$
   ♦ If $\small{f(2a - x)~=~-f(x)}$
   
Here, 2a = $\small{2 \pi}$

So $\small{f(2a-x) = \cos^5(2 \pi - x)=\left(\cos(2\pi -x) \right)^5 =\left(\cos(x) \right)^5 = \cos^5 x = f(x)}$

So we can apply P₆ (i). We get:

$\small{\int_{0}^{2\pi}{\left[\cos^5 x \right]dx}=2\int_{0}^{\pi}{\left[\cos^5 x \right]dx}~=~2I}$

2. We want to find I. We will apply:

$\bf{{\rm{P_4}}:\int_0^a{\left[f(x) \right]dx}~=~\int_0^a{\left[f(a-x) \right]dx}}$

We get:

$\small{I = \int_{0}^{\pi}{\left[\cos^5 x \right]dx}=\int_{0}^{\pi}{\left[\cos^5 (\pi - x) \right]dx} }$

$\small{\int_{0}^{\pi}{\left[(-\cos x)^5 \right]dx} =\int_{0}^{\pi}{\left[(-1)\cos^5x\right]dx}=(-1)\int_{0}^{\pi}{\left[\cos^5x\right]dx}=(-1)I}$

[$\small{\because~\cos(\pi - x)=-\cos x}$]

• Then $\small{2I = 0}$

3. Therefore, from the result in (1), we get:

$\small{\int_{0}^{2\pi}{\left[\cos^5 x \right]dx}~=~2I~=~0}$

Solved Example 23.116
Evaluate $\small{\int_{0}^{\frac{\pi}{2}}{\left[2 \log \sin(x) - \log \sin(2x) \right]dx}}$
Solution:
1. Let us rearrange the given function:

$\small{f(x)=2 \log \sin(x) - \log \sin(2x)= \frac{}{} \log \sin^2(x) - \log \sin(2x)}$

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

2. So we can write:

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

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

3. Applying P₄, we get:

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

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

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

4. Let us compare the results from (2) and (3):

• From (2), we have:

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

• From (3), we have:

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

5. Adding the two results, we get:

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

$\small{\Rightarrow I=(-1)\int_{0}^{\frac{\pi}{2}}{\left[\log 2\right]dx}=(-1)(\log 2)[x]_0^{\frac{\pi}{2}}}$

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

Solved Example 23.117
Evaluate $\small{\int_{0}^{\frac{\pi}{2}}{\left[\log \sin(x) \right]dx}}$
Solution:
1. The limits are $\small{0~\rm{and}~\frac{\pi}{2}}$. So we will apply:

$\bf{{\rm{P_4}}:\int_0^a{\left[f(x) \right]dx}~=~\int_0^a{\left[f(a-x) \right]dx}}$

We get:
$\small{I~=~\int_{0}^{\frac{\pi}{2}}{\left[\log \sin(x)\right]dx}~=~\int_{0}^{\frac{\pi}{2}}{\left[\log \sin \left(\frac{\pi}{2} - x \right) \right]dx}  }$

$\small{~=~\int_{0}^{\frac{\pi}{2}}{\left[\log \cos(x)\right]dx}}$

2. Now, I + I = 2I =

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

$\small{~=\int_{0}^{\frac{\pi}{2}}{\Big[\log \left[\sin(x)\,\cos x \right] \Big]dx}}$

3. Let us add and subtract "log 2". We get:

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

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

$\small{\Rightarrow 2I~=~\int_{0}^{\frac{\pi}{2}}{\Big[\log \left[\sin(2x) \right] \Big]dx}~-~\int_{0}^{\frac{\pi}{2}}{\Big[\log 2 \Big]dx}~=~I_1~-~I_2}$

4. Now we will evaluate $\small{I_1}$:

We have: $\small{I_1~=~\int_{0}^{\frac{\pi}{2}}{\Big[\log \left[\sin(2x) \right] \Big]dx}}$

• Put u = 2x

• Then $\small{\frac{du}{dx}~=~2 ~\Rightarrow 2dx = du}$

Also, When x approach zero, u approach zero

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

Then $\small{I_1~=~\int_{0}^{\frac{\pi}{2}}{\Big[\left(\frac{2}{2} \right)\log \left[\sin(2x) \right] \Big]dx}~=~\int_{0}^{\pi}{\Big[\left(\frac{1}{2} \right)\log \left[\sin(u) \right] \Big]du}}$

$\small{\Rightarrow I_1~=~\left(\frac{1}{2} \right) \int_{0}^{\pi}{\Big[\log \left[\sin(u) \right] \Big]du}}$ 

5. Let us apply:

$\bf{{\rm{P_6}}:}$

$\bf{\int_0^{2a}{\left[f(x) \right]dx}~=~2 \int_0^a{\left[f(x) \right]dx}}$
   ♦ If $\small{f(2a - x)~=~f(x)}$

$\bf{\int_0^{2a}{\left[f(x) \right]dx}~=~0}$
   ♦ If $\small{f(2a - x)~=~-f(x)}$
   
• Here assume that, 2a = $\small{\pi}$

• Given $\small{f(u)=\log[\sin(u)]}$

• $\small{f(2a-u)=\log[\sin(\pi-u)]~=~\log[\sin (u)]~=~f(u)}$

So we can apply P₆ (i). We get:

$\small{I_1~=~\left(\frac{1}{2} \right) \int_{0}^{\pi}{\Big[\log \left[\sin(u) \right] \Big]du}}$

$\small{\int_{0}^{\pi}{\Big[\log \left[\sin(u) \right] \Big]du}~=~2\int_{0}^{\frac{\pi}{2}}{\Big[\log \left[\sin(u) \right] \Big]du}}$

6. Then from (4), we get:

$\small{I_1~=~\left(\frac{1}{2} \right) \int_{0}^{\pi}{\Big[\log \left[\sin(u) \right] \Big]du}~=~\left(\frac{1}{2} \right)(2) \int_{0}^{\frac{\pi}{2}}{\Big[\log \left[\sin(u) \right] \Big]du}}$

$\small{\Rightarrow I_1~=~ \int_{0}^{\frac{\pi}{2}}{\Big[\log \left[\sin(u) \right] \Big]du}}$

7. Applying P₀ to the result in (6), we can write:

$\small{I_1~=~ \int_{0}^{\frac{\pi}{2}}{\Big[\log \left[\sin(x) \right] \Big]dx}}$

$\small{\Rightarrow I_1~=~ I}$

8. Now, based on the result in (3), we can write:

$\small{2I~=~I~-~I_2}$

$\small{\Rightarrow I~=~(-1)I_2}$

$\small{\Rightarrow I~=~(-1)\int_{0}^{\frac{\pi}{2}}{\Big[\log 2 \Big]dx}}$

$\small{\Rightarrow I~=~(-1){\Big[\log 2\left(\frac{\pi}{2} \right) \Big]dx}~=~\frac{-\pi\,\log 2}{2}}$

Solved Example 23.118
Evaluate $\small{\int_{0}^{\frac{\pi}{2}}{\left[\log\left(\frac{4 + 3 \sin x}{4 + 3 \cos x} \right) \right]dx}}$
Solution:
1. The limits are $\small{0~\rm{and}~a}$. So we will apply:

$\bf{{\rm{P_4}}:\int_0^a{\left[f(x) \right]dx}~=~\int_0^a{\left[f(a-x) \right]dx}}$

We get:

$\small{I~=~\int_{0}^{\frac{\pi}{2}}{\left[\log\left(\frac{4 + 3 \sin x}{4 + 3 \cos x} \right) \right]dx}~=~\int_{0}^{\frac{\pi}{2}}{\left[\log\left(4 + 3 \sin x \right)~-~\log\left(4 + 3 \cos x \right) \right]dx}  }$

$\small{~=~\int_{0}^{\frac{\pi}{2}}{\left[\log\left(4 + 3 \sin \left(\frac{\pi}{2} - x \right) \right)~-~\log\left(4 + 3 \cos \left(\frac{\pi}{2} - x \right) \right) \right]dx}  }$

$\small{~=~\int_{0}^{\frac{\pi}{2}}{\left[\log\left(4 + 3 \cos x  \right)~-~\log\left(4 + 3 \sin x \right) \right]dx}  }$

$\small{~=~(-1)\int_{0}^{\frac{\pi}{2}}{\left[\log\left(4 + 3 \sin x  \right)~-~\log\left(4 + 3 \cos x \right) \right]dx}  }$

$\small{~=~(-1)I }$

2. Therefore, 2I = 0, Which gives I = 0

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The link below gives a few more solved examples:

Exercise 23.11

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In the next section, we will see some miscellaneous solved examples.

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23.27 - Solved Examples on Properties of Definite Integrals

In the previous section, we saw some properties of definite integrals. We saw some solved examples also. In this section, we will see a few more solved examples.

First we will write the method to test whether a given function is odd, even, neither:

Even: $\small{f(-x) \,=\,f(x)}$ 

Odd: $\small{f(-x) \,=\,-f(x)}$ 

Neither: $\small{f(-x) \, \ne \,f(x)}$  and

              $\small{f(-x) \, \ne \,-f(x)}$

Let us see an example:

Check whether the function $\small{f(x)\,=\,x^2 + x}$ is even, odd or neither.

Solution:
1. $\small{f(-x)\,=\,(-x)^2 + (-x)\,=\,x^2 \,-\,x}$ 

2. $\small{-f(x)\,=\,-x^2\,-\,x}$

3. We see that:

$\small{f(-x) \, \ne \,f(x)}$ and  

$\small{f(-x) \, \ne \,-f(x)}$

So the given function is neither even nor odd

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Now we will proceed with the solved examples:

Solved Example 23.104
Evaluate $\small{\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}{\left[\sin^2 x \right]dx}}$
Solution:
1. The limits are $\small{\frac{-\pi}{4}~\rm{and}~\frac{\pi}{4}}$.
So we check whether $\small{f(x)=\sin^2 x}$ is even or odd.

(a) $\small{f(-x)\,=\,\sin^2(-x)\,=\,\left(\sin(-x) \right)^2}$ 

$\small{\,=\,\left(-\sin(x) \right)^2\,=\,\sin^2 x}$ 

$\small{\because \sin(-x) = -\sin x }$

(b) $\small{-f(x)\,=\,-\sin^2 x}$

(c) We see that:

$\small{f(-x) \,=\,f(x)}$

• So the given f(x) is an even function.

2. Since the given function is even, we can apply:

$\bf{{\rm{P_7~(i)}}:}$

$\bf{\int_{-a}^{a}{\left[f(x) \right]dx}~=~2 \int_0^a{\left[f(x) \right]dx}}$
   ♦ If $\small{f}$ is an even function

• We can write:

$\small{\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}{\left[\sin^2 x \right]dx}\,=\,2 \int_{0}^{\frac{\pi}{4}}{\left[\sin^2 x \right]dx}}$

3. Let $\small{I~=~\,2 \int_{0}^{\frac{\pi}{4}}{\left[\sin^2 x \right]dx}}$

• First we find the indefinite integral F.

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

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

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

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

4. Therefore,

$\small{I~=~F(\frac{\pi}{4})\,-\,F(0)~=~\left[x~-~ \frac{\sin(2x)}{2} \right]_{0}^{\frac{\pi}{4}}}$

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

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

Solved Example 23.105
Evaluate $\small{\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}{\left[\sin^2 x \right]dx}}$
Solution:
1. The limits are $\small{\frac{-\pi}{2}~\rm{and}~\frac{\pi}{2}}$.
So we check whether $\small{f(x)=\sin^2 x}$ is even or odd.

(a) $\small{f(-x)\,=\,\sin^2(-x)\,=\,\left(\sin(-x) \right)^2}$

$\small{\,=\,\left(-\sin(x) \right)^2\,=\,\sin^2 x}$

$\small{\because \sin(-x) = -\sin x }$

(b) $\small{-f(x)\,=\,-\sin^2 x}$

(c) We see that:

$\small{f(-x) \,=\,f(x)}$

• So the given f(x) is an even function.

2. Since the given function is even, we can apply:

$\bf{{\rm{P_7~(i)}}:}$

$\bf{\int_{-a}^{a}{\left[f(x) \right]dx}~=~2 \int_0^a{\left[f(x) \right]dx}}$
   ♦ If $\small{f}$ is an even function

• We can write:

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

3. Let $\small{I~=~\,2 \int_{0}^{\frac{\pi}{2}}{\left[\sin^2 x \right]dx}}$

• First we find the indefinite integral F.

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

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

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

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

4. Therefore,

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

$\small{~=~\left[\frac{\pi}{2}~-~ \frac{\sin(\pi)}{2} \right]~-~\left[0~-~ \frac{\sin(0)}{2} \right]}$

$\small{~=~\left[\frac{\pi}{2}~-~ \frac{0}{2} \right]~-~\left[0~-~ \frac{0}{2} \right]~=~\frac{\pi}{2}}$

Solved Example 23.106
Evaluate $\small{\int_{0}^{\frac{\pi}{2}}{\left[\cos^2 x \right]dx}}$
Solution:
1. The limits are $\small{0~\rm{and}~\frac{\pi}{2}}$.

$\small{\frac{\pi}{2}}$ is $\small{\left(2 \times\frac{\pi}{4} \right)}$

So we can apply:

$\bf{{\rm{P_5}}:\int_0^{2a}{\left[f(x) \right]dx}~=~\int_0^a{\left[f(x) \right]dx}~+~\int_0^a{\left[f(2a-x) \right]dx}}$

2. We get:

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

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

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

$\small{=\left[x \right]_{0}^{\frac{\pi}{4}}dx~=~\frac{\pi}{4}}$

Solved Example 23.107
Evaluate $\small{\int_{0}^{\pi}{\left[\frac{x \sin(x)}{1 + \cos^2 (x)} \right]dx}}$
Solution:
1. Here we can apply:

$\bf{{\rm{P_4}}:\int_0^a{\left[f(x) \right]dx}~=~\int_0^a{\left[f(a-x) \right]dx}}$

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

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

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

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

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

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

3. First we find the indefinite integral $\small{F~=~\frac{\pi}{2} \int{\left[\frac{\sin(x)}{1 + \cos^2(x)} \right]dx}}$.

• Put u = cos x

$\small{\Rightarrow \frac{du}{dx}=-\sin x \Rightarrow -\sin x \,dx = du}$

• Now F can be written as:

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

• This integration gives:

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

4. We wrote: u = cos x

When x approach the lower limit zero, u approach 1

When x approach the upper limit $\small{\pi}$, u approach −1.

5. So the given integral can be written as:

$\small{I~=~\frac{\pi}{2} \int_{0}^{\pi}{\left[\frac{\sin(x)}{1 + \cos^2(x)} \right]dx}=\frac{\pi}{2} \int_1^{-1}{\left[\frac{(-1)}{1 + u^2} \right]du}}$

$\small{~=(-1)\frac{\pi}{2} \int_{-1}^{1}{\left[\frac{(-1)}{1 + u^2} \right]du}~\rm{[by~P_1]}~=~\frac{\pi}{2} \int_{-1}^{1}{\left[\frac{1}{1 + u^2} \right]du}}$

$\small{~=~(2)\frac{\pi}{2} \int_{0}^{1}{\left[\frac{1}{1 + u^2} \right]du}~\rm{[by~P_7,~since ~this~is~even~function]}}$

$\small{~=~\pi \int_{0}^{1}{\left[\frac{1}{1 + u^2} \right]du}}$

6. Therefore,

$\small{I~=~F(1)\,-\,F(0)~=~\left[(\pi)\tan^{-1}u \right]_{0}^{1}}$

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

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

Solved Example 23.108
Evaluate $\small{\int_{-1}^{1}{\left[\sin^5 (x) \, \cos^4(x) \right]dx}}$
Solution:
1. The limits are −1 and 1.
So we check whether $\small{f(x)=\sin^5 (x) \, \cos^4(x)}$ is even or odd.

(a) $\small{f(-x)\,=\,\sin^5 (-x) \, \cos^4(-x)\,=\,\left(\sin(-x) \right)^5\,\left(\cos(-x) \right)^4}$

$\small{\,=\,\left(-\sin(x) \right)^5\,\left(\cos(x) \right)^4\,=\,-\sin^5 x \, \cos^4 x}$

$\small{\because \sin(-x) = -\sin x ~~\rm{and}~~\cos(-x) = \cos x}$

(b) $\small{-f(x)\,=\,-\sin^5 (x) \, \cos^4(x)}$

(c) We see that:

$\small{f(-x) \,=\,-f(x)}$

• So the given f(x) is an odd function.

2. Since the given function is odd, we can apply:

$\bf{{\rm{P_7~(ii)}}:}$

$\bf{\int_{-a}^{a}{\left[f(x) \right]dx}~=~0}$
   ♦ If $\small{f}$ is an odd function

• We can write: $\small{\int_{-1}^{1}{\left[\sin^5 (x) \, \cos^4(x) \right]dx}~=~0}$

Solved Example 23.109
Evaluate $\small{\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sin^4 x}{\sin^4 x + \cos^4 x} \right]dx}}$
Solution:
1. The limits are $\small{0~\rm{and}~\frac{\pi}{2}}$. So we will apply:

$\bf{{\rm{P_4}}:\int_0^a{\left[f(x) \right]dx}~=~\int_0^a{\left[f(a-x) \right]dx}}$

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

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

2. Now, I + I = 2I =

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

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

Therefore, $\small{I = \frac{\pi}{4}}$

Solved Example 23.110
Evaluate $\small{\int_{0}^{\frac{\pi}{2}}{\left[\frac{\cos^5 x}{\sin^5 x + \cos^5 x} \right]dx}}$
Solution:
1. The limits are $\small{0~\rm{and}~\frac{\pi}{2}}$. So we will apply:

$\bf{{\rm{P_4}}:\int_0^a{\left[f(x) \right]dx}~=~\int_0^a{\left[f(a-x) \right]dx}}$

We get:
$\small{I~=~\int_{0}^{\frac{\pi}{2}}{\left[\frac{\cos^5 x}{\sin^5 x + \cos^5 x} \right]dx}=\int_{0}^{\frac{\pi}{2}}{\left[\frac{\cos^5 \left(\frac{\pi}{2} - x \right)}{\sin^5 \left(\frac{\pi}{2} - x \right) + \cos^5 \left(\frac{\pi}{2} - x \right)} \right]dx}}$

$\small{~=~\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sin^5 x}{\cos^5 x + \sin^5 x} \right]dx}}$

2. Now, I + I = 2I =

$\small{\int_{0}^{\frac{\pi}{2}}{\left[\frac{\cos^5 x}{\sin^5 x + \cos^5 x} \right]dx}+\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sin^5 x}{\sin^5 x + \cos^5 x} \right]dx}=\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sin^5 x + \cos^5 x}{\sin^5 x + \cos^5 x} \right]dx}}$

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

Therefore, $\small{I = \frac{\pi}{4}}$

Solved Example 23.111
Evaluate $\small{\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} \right]dx}}$
Solution:
1.1. The limits are $\small{0~\rm{and}~\frac{\pi}{2}}$. So we will apply:

$\bf{{\rm{P_4}}:\int_0^a{\left[f(x) \right]dx}~=~\int_0^a{\left[f(a-x) \right]dx}}$

We get:
$\small{I~=~\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} \right]dx}=\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sin^{\frac{3}{2}} \left(\frac{\pi}{2} - x \right)}{\sin^{\frac{3}{2}} \left(\frac{\pi}{2} - x \right) + \cos^{\frac{3}{2}} \left(\frac{\pi}{2} - x \right)} \right]dx}}$

$\small{~=~\int_{0}^{\frac{\pi}{2}}{\left[\frac{\cos^{\frac{3}{2}} x}{\cos^{\frac{3}{2}} x + \sin^{\frac{3}{2}} x} \right]dx}}$

2. Now, I + I = 2I =

$\small{\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} \right]dx}+\int_{0}^{\frac{\pi}{2}}{\left[\frac{\cos^{\frac{3}{2}} x}{\cos^{\frac{3}{2}} x + \sin^{\frac{3}{2}} x} \right]dx}=\int_{0}^{\frac{\pi}{2}}{\left[\frac{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x + \cos^{\frac{3}{2}} x} \right]dx}}$

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

Therefore, $\small{I = \frac{\pi}{4}}$

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

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23.26 - Properties of Definite Integrals

In the previous section, we saw evaluation of definite integrals by substitution. We saw some solved examples also. In this section, we will see some properties of definite integrals. We have to learn about seven properties. They are named as P₁, P₂, P₃, . . .

$\bf{{\rm{P_0}}:\int_a^b{\left[f(x) \right]dx}~=~\int_a^b{\left[f(t) \right]dt}}$

Proof:

The function is the same. Upper and lower limits are also the same. We are changing only the variable. So the definite integral will be the same.

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$\bf{{\rm{P_1}}:\int_a^b{\left[f(x) \right]dx}~=~-\int_b^a{\left[f(x) \right]dx}}$

Proof can be written in 4 steps:

1. $\small{\int_a^b{\left[f(x) \right]dx}~=~F(b)\,-\,F(a)}$

2. $\small{\int_b^a{\left[f(x) \right]dx}~=~F(a)\,-\,F(b)~=~-\left[F(b)\,-\,F(a) \right]}$

3. From (2), we get:
$\small{-\int_b^a{\left[f(x) \right]dx}~=~F(b)\,-\,F(a) }$

4. Comparing the results in (1) and (3), we get:
$\small{\int_a^b{\left[f(x) \right]dx}~=~-\int_b^a{\left[f(t) \right]dx}}$

5. Note that:
$\small{\int_a^a{\left[f(x) \right]dx}~=~F(a)\,-\,F(a)~=~0}$

---

$\bf{{\rm{P_2}}:\int_a^b{\left[f(x) \right]dx}~=~\int_a^c{\left[f(x) \right]dx}~+~\int_c^b{\left[f(x) \right]dx}}$

Proof can be written in 3 steps:

1. $\small{\int_a^c{\left[f(x) \right]dx}~=~F(c)\,-\,F(a)}$

2. $\small{\int_c^b{\left[f(x) \right]dx}~=~F(b)\,-\,F(c)}$

3. Therefore:

$\small{\int_a^c{\left[f(x) \right]dx}~+~\int_c^b{\left[f(x) \right]dx}}$

$\small{~=~F(c)\,-\,F(a)~+~F(b)\,-\,F(c)}$

$\small{~=~F(b)\,-\,F(a)}$

$\small{~=~\int_a^b{\left[f(x) \right]dx}}$

---

$\bf{{\rm{P_3}}:\int_a^b{\left[f(x) \right]dx}~=~\int_a^b{\left[f(a+b-x) \right]dx}}$

Proof can be written in 3 steps:

1. Put $\small{t=a+b-x}$

$\small{\implies \frac{dt}{dx}~=~-1}$

$\small{\implies -dx~=~dt}$

2. Rearranging the limits:

   ♦ When x approach the lower limit 'a', t approach 'b'

   ♦ When x approach the upper limit 'b', t approach 'c'
   
3. So we can write:

$\small{\int_a^b{\left[f(a+b-x) \right]dx}~=~\int_a^b{\left[(-1)(-1)f(a+b-x) \right]dx}}$

$\small{~=~\int_b^a{\left[(-1)\,f(t) \right]dt}~=~(-1)\int_b^a{\left[\,f(t) \right]dt}}$

$\small{~=~\int_a^b{\left[\,f(t) \right]dt}}$ (by P₁)

$\small{~=~\int_a^b{\left[\,f(x) \right]dx}}$ (by P0q) 

---

$\bf{{\rm{P_4}}:\int_0^a{\left[f(x) \right]dx}~=~\int_0^a{\left[f(a-x) \right]dx}}$

Proof can be written in 4 steps:

1. Put $\small{t=a-x}$

$\small{\implies \frac{dt}{dx}~=~-1}$

$\small{\implies -dx~=~dt}$

2. Rearranging the limits:

   ♦ When x approach the lower limit zero, t approach 'a'

   ♦ When x approach the upper limit 'a', t approach zero
   
3. So we can write:

$\small{\int_a^b{\left[f(a-x) \right]dx}~=~\int_a^b{\left[(-1)(-1)f(a-x) \right]dx}}$

$\small{~=~\int_a^0{\left[(-1)\,f(t) \right]dt}~=~(-1)\int_a^0{\left[\,f(t) \right]dt}}$

$\small{~=~\int_0^a{\left[\,f(t) \right]dt}}$ (by P₁)

$\small{~=~\int_0^a {\left[\,f(x) \right]dx}}$ (by P0q)

4. Note that, P₄ is a particular case of P₃.

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$\bf{{\rm{P_5}}:\int_0^{2a}{\left[f(x) \right]dx}~=~\int_0^a{\left[f(x) \right]dx}~+~\int_0^a{\left[f(2a-x) \right]dx}}$

Proof can be written in 4 steps:

1. Using P₂, we can write:

$\small{\int_0^{2a}{\left[f(x) \right]dx}~=~\int_0^a{\left[f(x) \right]dx}~+~\int_a^{2a}{\left[f(x) \right]dx}}$

• We can denote this as:

$\small{\int_0^{2a}{\left[f(x) \right]dx}~=~I_1~+~I_2}$

2. Put $\small{t=2a-x}$

$\small{\implies \frac{dt}{dx}~=~-1}$

$\small{\implies -dx~=~dt}$

Also, $\small{x = 2a - t}$

3. Rearranging the limits of I₂:

   ♦ When x approach the lower limit 'a', t approach 'a'

   ♦ When x approach the upper limit '2a', t approach zero
   
So I₂ can be written as:

$\small{\int_a^{2a}{\left[f(x) \right]dx}~=~\int_a^0{\left[f(2a-t) \right]dx}~=~\int_a^0{\left[(-1)(-1)f(2a-t) \right]dx}}$

$\small{~=~\int_a^0{\left[(-1)\,f(2a - t) \right]dt}~=~(-1)\int_a^0{\left[\,f(2a - t) \right]dt}}$

$\small{~=~\int_0^a{\left[\,f(2a - t) \right]dt}}$ (by P₁)

$\small{~=~\int_0^a {\left[\,f(2a - x) \right]dx}}$ (by P0q)

4. So from step (1), we get:

$\small{\int_0^{2a}{\left[f(x) \right]dx}~=~\int_0^a{\left[f(x) \right]dx}~+~\int_0^a {\left[\,f(2a - x) \right]dx}}$

---

$\bf{{\rm{P_6}}:}$

$\bf{\int_0^{2a}{\left[f(x) \right]dx}~=~2 \int_0^a{\left[f(x) \right]dx}}$
   ♦ If $\small{f(2a - x)~=~f(x)}$

$\bf{\int_0^{2a}{\left[f(x) \right]dx}~=~0}$
   ♦ If $\small{f(2a - x)~=~-f(x)}$

Proof can be written in 3 steps:

1. Using P₅, we can write:

$\small{\int_0^{2a}{\left[f(x) \right]dx}~=~\int_0^a{\left[f(x) \right]dx}~+~\int_0^a {\left[\,f(2a - x) \right]dx}}$

2. If $\small{f(2a-x)~=~f(x)}$, (1) becomes:

$\small{\int_0^{2a}{\left[f(x) \right]dx}~=~\int_0^a{\left[f(x) \right]dx}~+~\int_0^a{\left[f(x) \right]dx}~=~2\int_0^a{\left[f(x) \right]dx}}$

3. If $\small{f(2a-x)~=~-f(x)}$, (1) becomes:

$\small{\int_0^{2a}{\left[f(x) \right]dx}~=~\int_0^a{\left[f(x) \right]dx}~-~\int_0^a{\left[f(x) \right]dx}~=~0}$

---

$\bf{{\rm{P_7}}:}$

$\bf{\int_{-a}^{a}{\left[f(x) \right]dx}~=~2 \int_0^a{\left[f(x) \right]dx}}$
   ♦ If $\small{f}$ is an even function
         ✰ That is., if $\small{f(-x)~=~f(x)}$

$\bf{\int_{-a}^{a}{\left[f(x) \right]dx}~=~0}$
   ♦ If $\small{f}$ is an odd function
         ✰ That is., if $\small{f(-x)~=~-f(x)}$

Proof can be written in 7 steps:

1. Using P₂, we can write:

$\small{\int_{-a}^{a}{\left[f(x) \right]dx}~=~\int_{-a}^0{\left[f(x) \right]dx}~+~\int_0^a {\left[\,f(x) \right]dx}}$

• We can denote this as:

$\small{\int_{-a}^{a}{\left[f(x) \right]dx}~=~I_1~+~I_2}$

2. Put $\small{t=-x}$ in I₁

$\small{\implies \frac{dt}{dx}~=~-1}$

$\small{\implies -dx~=~dt}$

Also, $\small{x = - t}$

3. Rearranging the limits of I₁:

   ♦ When x approach the lower limit '-a', t approach 'a'

   ♦ When x approach the upper limit zero, t approach zero
   
4. So I₁ can be written as:

$\small{\int_{-a}^0{\left[f(x) \right]dx}~=~\int_{a}^0{\left[f(-t) \right]dx}~=~\int_{a}^0{\left[(-1)(-1)f(-t) \right]dx}}$

$\small{~=~\int_{a}^0{\left[(-1)\,f(- t) \right]dt}~=~(-1)\int_{a}^0{\left[f(-t) \right]dt}}$

$\small{~=~(-1)\int_{a}^0{\left[f(-x) \right]dx}}$ (by P0q)

$\small{~=~\int_{0}^{a}{\left[f(-x) \right]dx}}$ (by P₁)

5. So from step (1), we get:

$\small{\int_{-a}^{a}{\left[f(x) \right]dx}~=~\int_{0}^a{\left[f(-x) \right]dx}~+~\int_0^a {\left[\,f(x) \right]dx}}$

6. If $\small{f}$ is an even function, then $\small{f(-x)~=~f(x)}$

• In such a situation, step (5) becomes:

$\small{\int_{-a}^{a}{\left[f(x) \right]dx}~=~\int_{0}^a{\left[f(x) \right]dx}~+~\int_0^a {\left[\,f(x) \right]dx}~=~2\int_0^a {\left[\,f(x) \right]dx}}$

7. If $\small{f}$ is an odd function, then $\small{f(-x)~=~-f(x)}$

• In such a situation, step (5) becomes:

$\small{\int_{-a}^{a}{\left[f(x) \right]dx}~=~-\int_{0}^a{\left[f(x) \right]dx}~+~\int_0^a {\left[\,f(x) \right]dx}~=~0}$

◼ We can cite two examples related to P₇:

Example 1:
This can be written in 5 steps:

1. $\small{f(x) \,=\,x^2}$ is an even function.

• Suppose that, we need to evaluate $\small{\int_{-1}^{1}{\left[x^2 \right]dx}}$

2. The usual approach is:

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

3. But by applying P₇, we can straight away write:

$\small{\int_{-1}^{1}{\left[x^2 \right]dx}\,=\,2\int_{0}^{1}{\left[x^2 \right]dx}}$

$\small{\,=\,2 \left[\frac{x^3}{3} \right]_{0}^1\,=\,2\left[\frac{1}{3}\,-\,\frac{0}{3} \right]\,=\,\frac{2}{3}}$

4. We get the same result in both (2) and (3).

5. In the fig.23.24 below,

   ♦ the area of the blue region from x = -1 to x = 0 is 1/3

   ♦ the area of the blue region from x = 0 to x = 1 is also 1/3

Fig.23.24

 

 

• Both areas are positive. So the total area from x = -1 to x = 1 will be twice (1/3)

Example 2:
This can be written in 5 steps:

1. $\small{f(x) \,=\,x^3}$ is an odd function.

• Suppose that, we need to evaluate $\small{\int_{-1}^{1}{\left[x^3 \right]dx}}$

2. The usual approach is:

$\small{\int_{-1}^{1}{\left[x^3 \right]dx}\,=\,\left[\frac{x^4}{4} \right]_{-1}^1\,=\,\frac{1}{4}\,-\,\frac{1}{4}\,=\,0}$

3. But by applying P₇, we can straight away write:

$\small{\int_{-1}^{1}{\left[x^3 \right]dx}\,=\,0}$

4. We get the same result in both (2) and (3).

5. In the fig.23.25 below,

   ♦ the area of the blue region from x = -1 to x = 0 is 1/4

   ♦ the area of the blue region from x = 0 to x = 1 is also 1/4

Fig.23.25

• The area from x = -1 to x = 0 is -ve. But the area from x = 0 to x = 1 is +ve. So the total area from x = -1 to x = 1 will be zero.

---

Now we will see some solved examples:

Solved Example 23.96
Given that:
$\small{\int_{1}^{5}{\left[f(x) \right]dx}\,=\,-3}$
$\small{\int_{2}^{5}{\left[f(x) \right]dx}\,=\,4}$

Find:
$\small{\int_{1}^{2}{\left[f(x) \right]dx}}$
Solution:
1. Applying P₂, we can write:

$\small{\int_{1}^{5}{\left[f(x) \right]dx}\,=\,\int_{1}^{2}{\left[f(x) \right]dx}\,+\,\int_{2}^{5}{\left[f(x) \right]dx}}$

2. Substituting the known values, we get:

$\small{-3\,=\,\int_{1}^{2}{\left[f(x) \right]dx}\,+\,4}$

3. Therefore,

$\small{\int_{1}^{2}{\left[f(x) \right]dx}\,=\,-3 - 4  = -7}$

Solved Example 23.97

Evaluate $\small{\int_{-2}^{3}{\left[|x| \right]dx}}$
Solution:
1. We know that:
$\left|x \right| = \begin{cases} x,  & \text{if}~x \ge 0 \\[1.5ex] -x, & \text{if}~x<0 \end{cases}$

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

$\small{\int_{-2}^{3}{\left[\left|x \right| \right]dx}\,=\,\int_{-2}^{0}{\left[\left|x \right| \right]dx}\,+\,\int_{0}^{3}{\left[\left|x \right| \right]dx}}$

• We will denote it as:

$\small{\int_{-2}^{3}{\left[\left|x \right| \right]dx}\,=\,I_1\,+\,I_2}$

3. Choosing the appropriate segments:

• For I₁, we must use the segment: $\small{\left|x \right|\,=\,-x,~\text{if}~x<0}$

• For I₂, we must use the segment: $\small{\left|x \right|\,=\,x,~\text{if}~x \ge 0}$

4. So from (2), we get:

$\small{\int_{-2}^{3}{\left[\left|x \right| \right]dx}\,=\,\int_{-2}^{0}{\left[-x \right]dx}\,+\,\int_{0}^{3}{\left[x \right]dx}}$

$\small{\,=\,(-1)\int_{-2}^{0}{\left[x \right]dx}\,+\,\int_{0}^{3}{\left[x \right]dx}}$

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

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

$\small{\,=\,(-1)\left[-2 \right]\,+\,\left[\frac{9}{2} \right]~=~\frac{13}{2}}$

Solved Example 23.98

Evaluate $\small{\int_{-1}^{1}{\left[1\,-\,|x| \right]dx}}$
Solution:
1. We know that:
$\left|x \right| = \begin{cases} x,  & \text{if}~x \ge 0 \\[1.5ex] -x, & \text{if}~x<0 \end{cases}$

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

$\small{\int_{-1}^{1}{\left[1\,-\,\left|x \right| \right]dx}\,=\,\int_{-1}^{0}{\left[1\,-\,\left|x \right| \right]dx}\,+\,\int_{0}^{1}{\left[1\,-\,\left|x \right| \right]dx}}$

• We will denote it as:

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

3. Choosing the appropriate segments:

• For I₁, we must use the segment: $\small{\left|x \right|\,=\,-x,~\text{if}~x<0}$

• For I₂, we must use the segment: $\small{\left|x \right|\,=\,x,~\text{if}~x \ge 0}$

4. So from (2), we get:

$\small{\int_{-1}^{1}{\left[1\,-\,\left|x \right| \right]dx}\,=\,\int_{-1}^{0}{\left[1+x \right]dx}\,+\,\int_{0}^{1}{\left[1\,-\,x \right]dx}}$

$\small{\,=\,\left[x\,+\,\frac{x^2}{2} \right]_{-1}^{0}\,+\,\left[x\,-\,\frac{x^2}{2} \right]_{0}^{1}}$

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

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

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

$\small{\,=\,\left[\frac{1}{2} \right]\,+\,\left[\frac{1}{2}\right]\,=\,1}$

Solved Example 23.99
Evaluate $\small{\int_{0}^{4}{\left[\left|x-1 \right| \right]dx}}$
Solution:
1. 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, $\small{\left|x-1 \right|\,=\,-(x-1)}$

2. 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, $\small{\left|x-1 \right|\,=\,x-1}$

3. 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, $\small{\left|x-1 \right|\,=\,x-1}$

4. Now we can write a piece wise function:

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

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

$\small{\int_{0}^{4}{\left[\left|x-1 \right| \right]dx}\,=\,\int_{0}^{1}{\left[\left|x-1 \right| \right]dx}\,+\,\int_{1}^{4}{\left[\left|x-1 \right| \right]dx}}$

• We will denote it as:

$\small{\int_{0}^{4}{\left[\left|x-1 \right| \right]dx}\,=\,I_1\,+\,I_2}$

6. Choosing the appropriate segments:

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

• For I₂, we must use the segment: $\small{\left|x-1 \right|\,=\,x-1,~\text{if}~x \ge 1}$

7. So from (5), we get:

$\small{\int_{0}^{4}{\left[\left|x-1 \right| \right]dx}\,=\,\int_{0}^{1}{\left[-(x-1) \right]dx}\,+\,\int_{1}^{4}{\left[x-1 \right]dx}}$

$\small{\,=\,(-1)\int_{0}^{1}{\left[x-1 \right]dx}\,+\,\int_{1}^{4}{\left[x-1 \right]dx}}$

$\small{\,=\,(-1)\left[\frac{x^2}{2}\,-\,x \right]_{0}^{1}\,+\,\left[\frac{x^2}{2}\,-\,x \right]_{1}^{4}}$

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

$\small{\,=\,(-1)\left[\frac{1}{2}\,-\,1\,-\,\left(0 \right) \right]\,+\,\left[\frac{16}{2}\,-\,4\,-\,\left(\frac{1}{2}\,-\,1 \right) \right]}$

$\small{\,=\,(-1)\left[-\frac{1}{2} \right]\,+\,\left[4\,+\,\frac{1}{2}\right]\,=\,5}$

Solved Example 23.100
Evaluate $\small{\int_{-1}^{2}{\left[\left|x^3\,-\,x \right| \right]dx}}$
Solution:
1. Let us determine the intervals where (x³ - x) is +ve or -ve.

2. For that, first we need to solve the equation: $\small{f(x)=x^3 - x = x(x^2 - 1)=0}$

• The solution is: x = −1, x = 0 and x = 1

• So the given interval [−1,2] can be split into three intervals: [−1,0], [0,1] and [1,2]  

3. Consider the interval (−1,0). In this interval, all x values are -ve fractions. So $\small{x^2}$ will be a +ve fraction. Consequently, $\small{(x^2 - 1)}$ will be -ve and $\small{x(x^2 - 1)}$ will be +ve

So for this interval, we can write:

$\small{\left|x(x^2 - 1) \right|=x(x^2 - 1)~\text{if}~-1 < x < 0}$

4. Consider the interval (0,1). In this interval, all x values are +ve fractions. So $\small{x^2}$ will be a +ve fraction. Consequently, $\small{(x^2 - 1)}$ will be -ve and $\small{x(x^2 - 1)}$ will be -ve

So for this interval, we can write:

$\small{\left|x(x^2 - 1) \right|=-x(x^2 - 1)~\text{if}~0 < x < 1}$

5. Consider the interval (1,2). In this interval, all x values are greater than 1. So $\small{x^2}$ will be greater than 1. Consequently, $\small{(x^2 - 1)}$ will be +ve and $\small{x(x^2 - 1)}$ will be +ve.

So for this interval, we can write:

$\small{\left|x(x^2 - 1) \right|=x(x^2 - 1)~\text{if}~1 < x < 2}$

6. Consider the exact points x = −1, x = 0, x = 1 and x = 2

• When x = −1, $\small{x(x^2 - 1)} = 0$

So for this point, we can write:

$\small{\left|x(x^2 - 1) \right|= \pm x(x^2 - 1)~\text{if}~ x = -1}$

• When x = 0, $\small{x(x^2 - 1)} = 0$

So for this point, we can write:

$\small{\left|x(x^2 - 1) \right|= \pm x(x^2 - 1)~\text{if}~ x = 0}$

• When x = 1, $\small{x(x^2 - 1)} = 0$

So for this point, we can write:

$\small{\left|x(x^2 - 1) \right|= \pm x(x^2 - 1)~\text{if}~ x = 1}$

• When x = 2, $\small{x(x^2 - 1)} = 6$

So for this point, we can write:

$\small{\left|x(x^2 - 1) \right|= +x(x^2 - 1)~\text{if}~ x = 2}$

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

$\left|x(x^2 - 1) \right| = \begin{cases} x(x^2 - 1),  & \text{if}~-1 \le x \le 0 \\[1.5ex] -x(x^2 - 1), & \text{if}~~0 \le x \le 1 \\[1.5ex] x(x^2 - 1), & \text{if}~~1 \le x \le 2  \end{cases}$

8. The integrand changes at −1, 0 and 1. So we will split the interval at those points. Then by applying P₂, it can be written as:

$\small{\int_{-1}^{2}{\left[\left|x^3-x \right| \right]dx}\,=\,\int_{-1}^{2}{\left[\left|x(x^2-1) \right| \right]dx}}$

$\small{\,=\,\int_{-1}^{0}{\left[\left|x(x^2-1) \right| \right]dx}\,+\,\int_{0}^{1}{\left[\left|x(x^2-1) \right| \right]dx}\,+\,\int_{1}^{2}{\left[\left|x(x^2-1) \right| \right]dx}}$

• We will denote it as:

$\small{\int_{-1}^{2}{\left[\left|x(x^2-1) \right| \right]dx}\,=\,I_1\,+\,I_2\,+\,I_3}$

9. Choosing the appropriate segments:

• For I₁, we must use the segment: $\small{\left|x(x^2-1) \right|\,=\,x(x^2-1),~\text{if}~-1 \le x \le 0}$

• For I₂, we must use the segment: $\small{\left|x(x^2-1) \right|\,=\,-x(x^2-1),~\text{if}~0 \le x \le 1}$

• For I₃, we must use the segment: $\small{\left|x(x^2-1) \right|\,=\,x(x^2-1),~\text{if}~1 \le x \le 2}$

10. So from (8), we get:

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

$\small{\,=\,\int_{-1}^{0}{\left[x(x^2-1) \right]dx}\,+\,\int_{0}^{1}{\left[-x(x^2-1) \right]dx}\,+\,\int_{1}^{2}{\left[x(x^2-1) \right]dx}}$

$\small{\,=\,\int_{-1}^{0}{\left[x^3-x \right]dx}\,+\,(-1)\int_{0}^{1}{\left[x^3-x \right]dx}\,+\,\int_{1}^{2}{\left[x^3-x \right]dx}}$

$\small{\,=\,\left[\frac{x^4}{4}\,-\,\frac{x^2}{2} \right]_{-1}^{0}\,+\,(-1) \left[\frac{x^4}{4}\,-\,\frac{x^2}{2} \right]_{0}^{1}\,+\, \left[\frac{x^4}{4}\,-\,\frac{x^2}{2} \right]_{1}^{2}}$

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

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

$\small{\,=\,\left[0\,-\,\left(\frac{-1}{4} \right) \right]\,+\,(-1)\left[\frac{-1}{4}\,-\,\left(0 \right) \right]+\left[2\,-\,\left(\frac{-1}{4} \right) \right]}$

$\small{\,=\,\left[\frac{1}{4} \right]\,+\,\left[\frac{1}{4} \right]+\left[\frac{9}{4} \right]\,=\,\frac{11}{4}}$

Solved Example 23.101
Evaluate $\small{\int_{-1}^{0}{\left[\left|4x\,+\,3 \right| \right]dx}}$
Solution:
1. Let us determine the intervals where (4x + 3) is +ve or -ve.

2. For that, first we need to solve the inequality: $\small{4x + 3 < 0}$

• The solution can be obtained as follows:
$\small{4x + 3 - 3  < 0 - 3}$
$\small{\Rightarrow 4x  <  - 3}$
$\small{\Rightarrow x  <  - \frac{3}{4}}$

That means, when x is less than $\small{- \frac{3}{4}}$, (4x+3) will be less than zero.  

3. Next  we need to solve the inequality: $\small{4x + 3 > 0}$

• The solution can be obtained as follows:
$\small{4x + 3 - 3  > 0 - 3}$
$\small{\Rightarrow 4x  >  - 3}$
$\small{\Rightarrow x  >  - \frac{3}{4}}$

That means, when x is greater than $\small{- \frac{3}{4}}$, (4x+3) will be greater than zero.

4. We have:

 $\small{\frac{-3}{4}\,=\,-0.75}$

• So the given interval [−1,0] can be split into two intervals:

$\small{\left(-1,-0.75 \right),~\left(-0.75,0 \right)}$

5. Consider the interval (−1, −0.75)
Based on (2), we can write:
(4x+3) will be −ve in this interval

So for this interval, we can write:

$\small{\left|4x + 3 \right|= -(4x + 3)~\text{if}~-1 < x < -0.75}$

6. Consider the interval (-0.75,0)
Based on (3), we can write:
(4x+3) will be +ve in this interval

So for this interval, we can write:
$\small{\left|4x + 3 \right|= (4x + 3)~\text{if}~-0.75 < x < 0}$

7. Consider the exact points x = −1, x = −0.75, and x = 0

• When x = −1, $\small{4x + 3} = -1$

So for this point, we can write:

$\small{\left|4x + 3 \right|= -(4x + 3)~\text{if}~ x = -1}$

• When x = -0.75, $\small{4x + 3} = 0$

So for this point, we can write:

$\small{\left|4x + 3 \right|= \pm(4x + 3)~\text{if}~ x = -0.75}$

• When x = 0, $\small{4x + 3} = 3$

So for this point, we can write:

$\small{\left|4x + 3 \right|= 4x + 3~\text{if}~ x = 0}$

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

$\left|4x + 3 \right| = \begin{cases} -(4x + 3),  & \text{if}~-1 \le x \le -0.75  \\[1.5ex] 4x + 3, & \text{if}~~-0.75 \le x \le 0  \end{cases}$

9. The integrand changes at -0.75. So we will split the interval at that point. Then by applying P₂, it can be written as:

$\small{\int_{-1}^{0}{\left[\left|4x+3 \right| \right]dx}\,=\,\int_{-1}^{-0.75}{\left[\left|4x+3 \right| \right]dx}\,+\,\int_{-0.75}^{0}{\left[\left|4x+3 \right| \right]dx}}$

• We will denote it as:

$\small{\int_{-1}^{0}{\left[\left|4x+3 \right| \right]dx}\,=\,I_1\,+\,I_2}$

10. Choosing the appropriate segments:

• For I₁, we must use the segment:

$\small{\left|4x + 3 \right|\,=\,-(4x+3),~\text{if}~-1 \le x \le -0.75}$

• For I₂, we must use the segment:

$\small{\left|4x + 3 \right|\,=\,4x+3,~\text{if}~-0.75 \le x \le 0}$

11. So from (9), we get:

$\small{\int_{-1}^{0}{\left[\left|4x+3 \right| \right]dx}=}$

$\small{\,=\,\int_{-1}^{-0.75}{\left[-(4x+3) \right]dx}\,+\,\int_{-0.75}^{1}{\left[4x+3 \right]dx}}$

$\small{\,=\,(-1)\int_{-1}^{-0.75}{\left[4x+3 \right]dx}\,+\,\int_{-0.75}^{1}{\left[4x+3 \right]dx}}$

$\small{\,=\,(-1)\left[\frac{4 x^2}{2}\,+\,3x \right]_{-1}^{-0.75}\,+\,\left[\frac{4 x^2}{2}\,+\,3x \right]_{-0.75}^{0}}$

$\small{\,=\,(-1)\left[2x^2\,+\,3x \right]_{-1}^{-0.75}\,+\,\left[2x^2\,+\,3x \right]_{-0.75}^{0}}$

$\small{\,=\,(-1)\left[2(-0.75)^2\,+\,3(-0.75)\,-\,\left(2(-1)^2\,+\,3(-1) \right) \right]}$

$\small{~~~~~~+\left[2(0)^2\,+\,3(0)\,-\,\left(2(-0.75)^2\,+\,3(-0.75) \right) \right]}$

$\small{\,=\,\frac{5}{4}}$

Solved Example 23.102
Evaluate $\small{\int_{-5}^{5}{\left[\left|x+2 \right| \right]dx}}$
Solution:
1. 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 −2, (x+2) will be -ve.

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

2. 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, $\small{\left|x+2 \right|\,=\,x+2}$

3. 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, $\small{\left|x+2 \right|\,=\,x+2}$

4. Now we can write a piece wise function:

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

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

$\small{\int_{-5}^{5}{\left[\left|x+2 \right| \right]dx}\,=\,\int_{-5}^{-2}{\left[\left|x+2 \right| \right]dx}\,+\,\int_{-2}^{5}{\left[\left|x+2 \right| \right]dx}}$

• We will denote it as:

$\small{\int_{-5}^{5}{\left[\left|x+2 \right| \right]dx}\,=\,I_1\,+\,I_2}$

6. Choosing the appropriate segments:

• For I₁, we must use the segment: $\small{\left|x+2 \right|\,=\,-(x+2),~\text{if}~x<-2}$

• For I₂, we must use the segment: $\small{\left|x+2 \right|\,=\,x+2,~\text{if}~x \ge -2}$

7. So from (5), we get:

$\small{\int_{-5}^{5}{\left[\left|x+2 \right| \right]dx}\,=\,\int_{-5}^{-2}{\left[-(x+2) \right]dx}\,+\,\int_{-2}^{5}{\left[x+2 \right]dx}}$

$\small{\,=\,(-1)\int_{-5}^{-2}{\left[x+2 \right]dx}\,+\,\int_{-2}^{5}{\left[x+2 \right]dx}}$

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

$\small{\,=\,(-1)\left[\frac{(-2)^2}{2}\,+\,2(-2)\,-\,\left(\frac{(-5)^2}{2}\,+\,2(-5) \right) \right]\,+\,\left[\frac{5^2}{2}\,+\,10\,-\,\left(\frac{(-2)^2}{2}\,+\,2(-2) \right) \right]}$

$\small{\,=\,(-1)\left[2\,-\,4\,-\,\left(\frac{25}{2}\,-\,10 \right) \right]\,+\,\left[\frac{25}{2}\,+\,10\,-\,\left(2\,-\,4 \right) \right]}$

$\small{\,=\,(-1)\left[-2\,-\,\left(\frac{25}{2}\,-\,10 \right) \right]\,+\,\left[\frac{25}{2}\,+\,10\,+\,2 \right]}$

$\small{\,=\,2+\frac{25}{2}-10+\frac{25}{2}+12~=~29}$

Solved Example 23.103
Evaluate $\small{\int_{2}^{8}{\left[\left|x-5 \right| \right]dx}}$
Solution:
1. Let us determine the intervals where (x−5) is +ve or −ve. For that, first we solve the inequality: x−5<0

We have: $\small{x-5 < 0}$

$\small{\Rightarrow x < 5}$

• So when x is less than 5, (x−5) will be -ve.

• That means, when x is less than 5, $\small{\left|x-5 \right|\,=\,-(x-5)}$

2. Next we solve the inequality:x−5>0

We have: $\small{x-5 > 0}$

$\small{\Rightarrow x > 5}$

• So when x is greater than 5, (x-5) will be +ve.

• That means, when x is greater  than 5, $\small{\left|x-5 \right|\,=\,x-5}$

3. Also, we must solve the equation x-5 = 0

This gives x = 5

• So when x is equal to 5, (x-5) will be zero.

• That means, when x is equal to 5, $\small{\left|x-5 \right|\,=\,x-5}$

4. Now we can write a piece wise function:

$\left|x-5 \right| = \begin{cases} x-5,  & \text{if}~x \ge 5 \\[1.5ex] -(x-5), & \text{if}~x<5  \end{cases}$

5. The integrand changes at 5. So we will split the interval at 5. Then by applying P₂, it can be written as:

$\small{\int_{2}^{8}{\left[\left|x-5 \right| \right]dx}\,=\,\int_{2}^{8}{\left[\left|x-5 \right| \right]dx}\,+\,\int_{2}^{8}{\left[\left|x-5 \right| \right]dx}}$

• We will denote it as:

$\small{\int_{2}^{8}{\left[\left|x-5 \right| \right]dx}\,=\,I_1\,+\,I_2}$

6. Choosing the appropriate segments:

• For I₁, we must use the segment: $\small{\left|x-5 \right|\,=\,-(x-5),~\text{if}~x<5}$

• For I₂, we must use the segment: $\small{\left|x-5 \right|\,=\,x-5,~\text{if}~x \ge 5}$

7. So from (5), we get:

$\small{\int_{2}^{8}{\left[\left|x-5 \right| \right]dx}\,=\,\int_{2}^{5}{\left[-(x-5) \right]dx}\,+\,\int_{5}^{8}{\left[x-5 \right]dx}}$

$\small{\,=\,(-1)\int_{2}^{5}{\left[x-5 \right]dx}\,+\,\int_{5}^{8}{\left[x-5 \right]dx}}$

$\small{\,=\,(-1)\left[\frac{x^2}{2}\,-\,5x \right]_{2}^{5}\,+\,\left[\frac{x^2}{2}\,-\,5x \right]_{5}^{8}}$

$\small{\,=\,(-1)\left[\frac{5^2}{2}\,-\,25\,-\,\left(\frac{2^2}{2}\,-\,10 \right) \right]\,+\,\left[\frac{8^2}{2}\,-\,40\,-\,\left(\frac{5^2}{2}\,-\,25 \right) \right]}$

$\small{\,=\,(-1)\left[\frac{25}{2}\,-\,25\,-\,\left(2\,-\,10 \right) \right]\,+\,\left[32\,-\,40\,-\,\left(\frac{25}{2}\,-\,25 \right) \right]}$

$\small{\,=\,(-1)\left[\frac{25}{2}\,-\,25\,-\,2\,+\,10  \right]\,+\,\left[32\,-\,40\,-\,\frac{25}{2}\,+\,25  \right]}$

$\small{\,=\,(-1)\left[\frac{25}{2}\,-\,17  \right]\,+\,\left[17-\,\frac{25}{2}  \right]}$

$\small{\,=\,\frac{-25}{2}+17+17-\frac{25}{2}~=~-25+34~=~9}$

---

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

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23.25 - Evaluation of Definite Integrals by Substitution

In the previous section, we saw how the fundamental theorem of calculus can be used to find the definite integral. We saw some solved examples also. In this section, we will see how the method of substitution can be used to speed up the process. We will demonstrate the method using a solved example.

Solved Example 23.88
Evaluate the definite integral $\small{\int_{1}^{2}{\left[5x^4 \sqrt{x^5 + 1} \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{-1}^{1}{\left[5x^4 \sqrt{x^5 + 1} \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[5x^4 \sqrt{x^5 + 1} \right]dx}}$

(a) Put $\small{u = (x^5 + 1)~\Rightarrow \frac{du}{dx} = 5x^4 \Rightarrow 5x^4 dx = du}$

(b) $\small{F~=~\int{\left[5x^4 \sqrt{x^5 + 1} \right]dx}~=~\int{\left[\sqrt{u} \right]du}}$

$\small{~=~\frac{u^{3/2}}{3/2}~=~\frac{2}{3} \left(x^5 + 1 \right)^{\frac{3}{2}}}$

2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(1)\,-\,F(-1)~=~\left[\frac{2}{3} \left(x^5 + 1 \right)^{\frac{3}{2}} \right]_{-1}^{1}}$

$\small{~=~\left[\frac{2}{3} \left(1^5 + 1 \right)^{\frac{3}{2}} \right]~-~\left[\frac{2}{3} \left((-1)^5 + 1 \right)^{\frac{3}{2}} \right]}$

$\small{~=~\left[\frac{2}{3} \left(2 \right)^{\frac{3}{2}} \right]~-~\left[\frac{2}{3} \left(0 \right)^{\frac{3}{2}} \right]}$

$\small{~=~\left[\frac{4 \sqrt 2}{3} \left(2 \right)^{\frac{3}{2}} \right]~-~\left[0 \right]~=~\frac{4 \sqrt 2}{3}}$

• The above two steps are already familiar to us. Let us see a method to speed up the process. For that, we will analyze the upper and lower limits.

• We wrote: $\small{u = x^5 + 1}$.
So when x gets closer and closer to -1, u gets closer and closer to zero.
Also, when x gets closer and closer to 1, u gets closer and closer to 2.

• So when using u as the variable, the lower and upper limits are zero and 2 respectively.

• We wrote: $\small{F~=~\int{\left[\sqrt{u} \right]du}~=~\frac{u^{3/2}}{3/2}~=~\frac{2}{3}u^{\frac{3}{2}}}$

• So we can write:

$\small{I~=~F(2)\,-\,F(0)~=~\left[\frac{2}{3}u^{\frac{3}{2}} \right]_{0}^{2}}$

$\small{~=~\left[\frac{2}{3} \left(2 \right)^{\frac{3}{2}} \right]~-~\left[\frac{2}{3} \left(0 \right)^{\frac{3}{2}} \right]}$

$\small{~=~\frac{4 \sqrt 2}{3}}$

---

Let us see a few more solved examples:

Solved Example 23.89
Evaluate the definite integral $\small{\int_{0}^{1}{\left[\frac{x}{x^2 + 1} \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{0}^{1}{\left[\frac{x}{x^2 + 1} \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[\frac{x}{x^2+1} \right]dx}}$

Put $\small{u~=~x^2 + 1 \Rightarrow \frac{du}{dx}~=~2x \Rightarrow 2x\,dx~=~du}$

So we have: $\small{F~=~\int{\left[\frac{2x}{2(x^2+1)} \right]dx}~=~\int{\left[\frac{1}{2(u)} \right]du}~=~\frac{\log \left|u \right|}{2}}$

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

   ♦ When x approach the lower limit zero, u approach 1

   ♦ When x approach the upper limit 1, u approach 2

3. So the given integral can be written as:

$\small{I~=~\int_{0}^{1}{\left[\frac{x}{x^2 + 1} \right]dx}~=~\int_1^2{\left[\frac{1}{2(u)} \right]du}}$

4. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(2)\,-\,F(1)~=~\left[\frac{\log \left|u \right|}{2} \right]_{1}^{2}}$

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

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

Solved Example 23.90
Evaluate the definite integral $\small{\int_{0}^{\frac{\pi}{2}}{\left[\sqrt{\sin \phi} \cos^5 \phi \right]d \phi}}$
Solution:
1. Let $\small{I~=~\int_{0}^{\frac{\pi}{2}}{\left[\sqrt{\sin \phi} \cos^5 \phi \right]d \phi}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[\sqrt{\sin \phi} \cos^5 \phi \right]d \phi}}$

$\small{\Rightarrow F~=~\int{\left[\sqrt{\sin \phi}\, \cos \phi\,\left(\cos^2 \phi \right)^2 \right]d \phi}}$

$\small{\Rightarrow F~=~\int{\left[\sqrt{\sin \phi}\, \cos \phi\,\left(1 - \sin^2 \phi \right)^2 \right]d \phi}}$

Put $\small{u~=~\sin \phi \Rightarrow \frac{du}{d\phi}~=~\cos \phi \Rightarrow \cos \phi\,d\phi~=~du}$

So we have: $\small{F~=~\int{\left[\sqrt{\sin \phi}\, \cos \phi\,\left(1 - \sin^2 \phi \right)^2 \right]d\phi}~=~\int{\left[\sqrt{u}\,\left(1 - u^2 \right)^2 \right]du}}$

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

$\small{~=~\frac{u^{3/2}}{3/2}~-~\frac{2 u^{7/2}}{7/2}~+~\frac{u^{11/2}}{11/2}~=~\frac{2 u^{3/2}}{3}~-~\frac{4 u^{7/2}}{7}~+~\frac{2 u^{11/2}}{11}}$

2. We wrote: $\small{u~=~\sin \phi}$

   ♦ When $\small{\phi}$ approach the lower limit zero, u approach 0

   ♦ When $\small{\phi}$ approach the upper limit $\small{\frac{\pi}{2}}$, u approach 1

3. So the given integral can be written as:

$\small{I~=~\int_{0}^{1}{\left[u^{1/2}~-~2 u^{5/2}~+~u^{9/2} \right]d \phi}}$

4. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(1)\,-\,F(0)~=~\left[\frac{2 u^{3/2}}{3}~-~\frac{4 u^{7/2}}{7}~+~\frac{2 u^{11/2}}{11} \right]_{0}^{1}}$

$\small{~=~\left[\frac{2 }{3}~-~\frac{4}{7}~+~\frac{2 }{11} \right]~-~\left[0 \right]}$

$\small{~=~\frac{64}{231} }$

Solved Example 23.91
Evaluate the definite integral $\small{\int_{0}^{1}{\left[\frac{\tan^{-1}x}{1+x^2} \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{0}^{1}{\left[\frac{\tan^{-1}x}{1+x^2} \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[\frac{\tan^{-1}x}{1+x^2} \right]dx}}$

Put $\small{u~=~\tan^{-1}x \Rightarrow \frac{du}{dx}~=~\frac{1}{1+x^2} \Rightarrow \frac{dx}{1+x^2}~=~du}$

So we have: $\small{F~=~\int{\left[u \right]du}~=~\frac{u^2}{2}}$

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

   ♦ When $\small{x}$ approach the lower limit zero, u approach 0

   ♦ When $\small{x}$ approach the upper limit $\small{1}$, u approach $\small{\frac{\pi}{4}}$

3. So the given integral can be written as:

$\small{I~=~\int_{0}^{\frac{\pi}{4}}{\left[u \right]du}}$

4. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(\frac{\pi}{4})\,-\,F(0)~=~\left[\frac{u^2}{2} \right]_{0}^{\frac{\pi}{4}}}$

$\small{~=~\left[\frac{\pi^2 }{32} \right]~-~\left[0 \right]~=~\frac{\pi^2 }{32}}$

Solved Example 23.92
Evaluate the definite integral $\small{\int_{0}^{1}{\left[\sin^{-1}\left(\frac{2x}{1+x^2} \right) \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{0}^{1}{\left[\sin^{-1}\left(\frac{2x}{1+x^2} \right) \right]dx}}$  

• First we find the indefinite integral F.

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

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

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

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

So we have:

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

(See solved example 23.54 of section 23.17)

$\small{~=~2x \tan^{-1}x~+~\log \left|1+x^2 \right|}$

2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(1)\,-\,F(0)~=~\left[2x \tan^{-1}x~+~\log \left|1+x^2 \right| \right]_{0}^{1}}$

$\small{~=~\left[2(1) \tan^{-1}1~+~\log \left|1+1^2 \right| \right]~-~\left[2(0) \tan^{-1}(0)~+~\log \left|1+(0)^2 \right| \right]~=~\frac{\pi^2 }{32}}$

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

Solved Example 23.93
Evaluate the definite integral $\small{\int_{0}^{2}{\left[x \sqrt{x+2} \right]dx}}$ (Put x+2 =  t²)
Solution:
1. Let $\small{I~=~\int_{0}^{2}{\left[x \sqrt{x+2} \right]dx}}$

Put $\small{x+2} =  t^2\Rightarrow x= t^2 - 2\Rightarrow\frac{dx}{dt}~=~2t \Rightarrow dx = 2t\, dt$

   ♦ When x approaches zero, t approaches $\small{\sqrt 2}$

   ♦ When x approaches 2, t approaches 2

So we get: $\small{F ~=~ \int_{\sqrt 2}^{2}{\left[(t^2 - 2) \sqrt{t^2} \right]2t\,dt}~=~ \int_{\sqrt 2}^{2}{\left[(t^2 - 2) t \right]2t\,dt}}$

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

2. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(2)\,-\,F(\sqrt 2)~=~\left[\frac{2 t^5}{5}~-~\frac{4t^3}{3} \right]_{\sqrt 2}^{2}}$

$\small{~=~\left[\frac{2 (2)^5}{5}~-~\frac{4(2)^3}{3} \right]~-~\left[\frac{2 (\sqrt 2)^5}{5}~-~\frac{4(\sqrt 2)^3}{3} \right]}$

$\small{~=~\left[\frac{2^6}{5}~-~\frac{2^5}{3} \right]~-~\left[\frac{2^3 (\sqrt 2)}{5}~-~\frac{2^3(\sqrt 2)}{3} \right]}$

$\small{~=~\frac{2^6}{5}~-~\frac{2^5}{3} ~-~\frac{2^3 (\sqrt 2)}{5}~+~\frac{2^3(\sqrt 2)}{3} }$

$\small{~=~\frac{(3)2^6~-~(5)2^5~-~(3)2^3(\sqrt 2)~+~(5)2^3(\sqrt 2)}{15}}$

$\small{~=~\frac{(3)2^6~-~(5)2^5~+~(2)2^3(\sqrt 2)}{15}~=~\frac{(3)2^6~-~(5)2^5~+~2^4(\sqrt 2)}{15}}$

$\small{~=~\frac{2^5 \left[6~-~5 \right]~+~2^4(\sqrt 2)}{15}~=~\frac{2^5 ~+~2^4(\sqrt 2)}{15}~=~\frac{2^4 \left[2 ~+~\sqrt 2 \right]}{15}}$

$\small{~=~\frac{16 \sqrt 2 \left[\sqrt 2 ~+~1 \right]}{15}}$

Solved Example 23.94
Evaluate the definite integral $\small{\int_{0}^{2}{\left[\frac{1}{x+4 - x^2} \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{0}^{2}{\left[\frac{1}{x+4 - x^2} \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[\frac{1}{x+4 - x^2} \right]dx}}$

• In our present case, a = -1, b = 1 and c = 4

2. So we can calculate u and $\small{k^2}$:

• $\small{u\,=\,x\,+\,\frac{b}{2a}~=~x\,+\,\frac{1}{2(-1)}~=~(x-\frac{1}{2})}$

• $\small{\pm k^2\,=\,\frac{c}{a}\,-\,\frac{b^2}{4a^2}~=~\frac{4}{-1}\,-\,\frac{(1)^2}{4(-1)^2}~=~-4\,-\,\frac{1}{4}~=~\frac{-17}{4}}$

3. So we want:

$\small{\int{\left[\frac{dx}{x+4 - x^2} \right]}~=~\frac{1}{a}\int{\left[\frac{du}{u^2~\pm k^2} \right]}}$

$\small{~=~\frac{1}{-1}\int{\left[\frac{dx}{(x-\frac{1}{2})^2~-~\frac{17}{4}} \right]}~=~(-1)\int{\left[\frac{dx}{(x-\frac{1}{2})^2~-~\frac{17}{4}} \right]}}$

[Recall that, we put u = x + b/(2a). So du = dx]

4. That means, we want the definite integral:

$\small{~(-1)\int_0^2{\left[\frac{dx}{(x-\frac{1}{2})^2~-~\frac{17}{4}} \right]}}$

(i) Put t = (x−1/2). Then dt/dx = 1, which gives dt = dx

   ♦ Also, when x approaches zero, t approaches -(1/2)

   ♦ Similarly, when x approaches 2, t approaches (3/2)

• So we want:

$\small{~(-1)\int_0^2{\left[\frac{dx}{(x-\frac{1}{2})^2~-~\frac{17}{4}} \right]}~=~~(-1)\int_{-1/2}^{3/2}{\left[\frac{dt}{(t)^2~-~\left(\frac{\sqrt{17}}{2} \right)^2} \right]}}$

(ii) We have formula: $\small{\int{\left[\frac{dt}{t^2\,-\,m^2} \right]}\,=\, \frac{1}{2m} \log \left| \frac{t-m}{t+m}  \right|}$

• In our present case, m = $\small{\frac{\sqrt{17}}{2}}$

5. So we get:
$\small{(-1)\int{\left[\frac{dt}{(t)^2~-~\left(\frac{\sqrt{17}}{2} \right)^2}  \right]}\,=\, \frac{-1}{\sqrt{17}} \log \left| \frac{t-\frac{\sqrt{17}}{2}}{t+\frac{\sqrt{17}}{2}}  \right|}$

6. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(-1/2)\,-\,F(3/2)~=~\left[\frac{-1}{\sqrt{17}} \log \left| \frac{t-\frac{\sqrt{17}}{2}}{t+\frac{\sqrt{17}}{2}}  \right| \right]_{-1/2}^{3/2}}$

$\small{~=~\left[\frac{-1}{\sqrt{17}} \log \left| \frac{\frac{3}{2}-\frac{\sqrt{17}}{2}}{\frac{3}{2}+\frac{\sqrt{17}}{2}}  \right| \right]~-~\left[\frac{-1}{\sqrt{17}} \log \left| \frac{\frac{-1}{2}-\frac{\sqrt{17}}{2}}{\frac{-1}{2}+\frac{\sqrt{17}}{2}}  \right|\right]}$

$\small{~=~\left[\frac{-1}{\sqrt{17}} \log \left|\frac{3 - \sqrt{17}}{3 + \sqrt{17}} \right| \right]~+~\left[\frac{1}{\sqrt{17}} \log \left|\frac{-1 - \sqrt{17}}{-1 + \sqrt{17}}  \right|\right]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\left[\log \left|\frac{-1 - \sqrt{17}}{-1 + \sqrt{17}}  \right|~-~ \log \left|\frac{3 - \sqrt{17}}{3 + \sqrt{17}} \right| \right]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\left[\log \left|\frac{(-1)(1 + \sqrt{17})}{(-1)(1 - \sqrt{17})}  \right|~-~ \log \left|\frac{3 - \sqrt{17}}{3 + \sqrt{17}} \right| \right]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\left[\log \left|\frac{1 + \sqrt{17}}{1 - \sqrt{17}}  \right|~-~ \log \left|\frac{3 - \sqrt{17}}{3 + \sqrt{17}} \right| \right]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\big[\log\left[(1 + \sqrt{17})(3 + \sqrt{17}) \right]~-~\log\left[(1 - \sqrt{17})(3 - \sqrt{17}) \right] \big]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\big[\log\left[20 + 4 \sqrt {17} \right]~-~\log\left[20 - 4 \sqrt{17} \right] \big]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\bigg[\log \left[\frac{20 + 4\sqrt{17}}{20 - 4\sqrt{17}} \right]  \bigg]~=~\frac{1}{\sqrt{17}}\,\bigg[\log \left[\frac{4(5 + \sqrt{17})}{4(5 - \sqrt{17})} \right]  \bigg]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\bigg[\log \left[\frac{5 + \sqrt{17}}{5 - \sqrt{17}} \right]  \bigg]~=~\frac{1}{\sqrt{17}}\,\bigg[\log \left[\frac{(5 + \sqrt{17})(5 + \sqrt{17})}{(5 - \sqrt{17})(5 + \sqrt{17})} \right]  \bigg]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\bigg[\log \left[\frac{25 + 10 \sqrt{17} + 17}{25 - 17} \right]  \bigg]~=~\frac{1}{\sqrt{17}}\,\bigg[\log \left[\frac{42 + 10 \sqrt{17}}{8} \right]  \bigg]}$

$\small{~=~\frac{1}{\sqrt{17}}\,\bigg[\log \left[\frac{21 + 5 \sqrt{17}}{4} \right]  \bigg]}$

Solved Example 23.95
Evaluate the definite integral $\small{\int_{-1}^{1}{\left[\frac{1}{x^2 + 2x + 5} \right]dx}}$
Solution:
1. Let $\small{I~=~\int_{-1}^{1}{\left[\frac{1}{x^2 + 2x + 5} \right]dx}}$

• First we find the indefinite integral F.

We have:
$\small{F~=~\int{\left[\frac{1}{x^2 + 2x + 5} \right]dx}}$

• In our present case, a = 1, b = 2 and c = 5

2. So we can calculate u and $\small{k^2}$:

• $\small{u\,=\,x\,+\,\frac{b}{2a}~=~x\,+\,1}$

• $\small{\pm k^2\,=\,\frac{c}{a}\,-\,\frac{b^2}{4a^2}~=~\frac{5}{1}\,-\,\frac{(2)^2}{4(1)^2}~=~5\,-\,1~=~4}$

3. So we want:

$\small{\int{\left[\frac{dx}{x^2 + 2x + 5} \right]}~=~\frac{1}{a}\int{\left[\frac{du}{u^2~\pm k^2} \right]}}$

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

[Recall that, we put u = x + b/(2a). So du = dx]

4. That means, we want the definite integral:

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

(i) Put t = (x+1). Then dt/dx = 1, which gives dt = dx

   ♦ Also, when x approaches -1, t approaches zero

   ♦ Similarly, when x approaches 1, t approaches 2

• So we want:

$\small{\int_{-1}^1{\left[\frac{dx}{(x+1)^2~+~2^2} \right]}~=~~\int_{0}^{2}{\left[\frac{dt}{(t)^2~+~\left(2 \right)^2} \right]}}$

(ii) We have formula: $\small{\int{\left[\frac{dt}{t^2\,+\,m^2} \right]}\,=\, \frac{1}{m} \tan^{-1}\left( \frac{t}{m} \right)}$

• In our present case, m = 2

5. So we get:
$\small{\int{\left[\frac{dt}{(t)^2~+~\left(2 \right)^2}  \right]}\,=\, \frac{1}{2} \tan^{-1}\left( \frac{t}{2} \right)}$

6. Therefore, by the second fundamental theorem of calculus, we get:
$\small{I~=~F(2)\,-\,F(0)~=~\left[\frac{1}{2} \tan^{-1}\left( \frac{t}{2} \right) \right]_{0}^{2}}$

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

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

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

$\small{~=~\frac{\pi}{8}}$

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The link below gives a few more solved examples:

Exercise 23.10

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

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