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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