In the previous section, we saw some miscellaneous examples on integrals. In this section, we will see a few more miscellaneous examples.
Solved Example 23.148
Integrate $\small{\frac{e^x}{(1+e^x)(2+e^x)}}$
Solution:
1. Let us put $\small{u = 1+e^x}$
• Then $\small{\frac{du}{dx}~=~e^x \Rightarrow e^x\,dx~=~du}$
• Also, since $\small{u = 1+e^x}$, we can write:
$\small{2+e^x = u+1}$
2. So we want:
$\small{I = \int{\left[\frac{e^x}{(1+e^x)(2+e^x)}\right]dx}= \int{\left[\frac{1}{(u)(u+1)}\right]du}= \int{\left[\frac{1}{u^2 + u}\right]du}}$
3. This is a standard integral of the form:
$\small{ \int{\left[\frac{px + q}{ax^2 + bx + c}\right]du}}$, where p = 0, q = 1, a = 1, b = 1 and c = 0
We get: $\small{I~=~\int{\left[\frac{1}{u^2 + u}\right]du}~=~(-1)\log \left| \frac{1}{u}~+~1 \right|~+~\rm{C}}$
$\small{~=~(-1)\log \left| \frac{1+u}{u} \right|~+~\rm{C}~=~\log \left| \frac{u}{1+u} \right|~+~\rm{C}}$
4. Substituting for u, we get:
$\small{I~=~\log \left| \frac{1+e^x}{2+e^x} \right|~+~\rm{C}}$
Solved Example 23.149
Integrate $\small{f'(ax+b) \left[f(ax+b) \right]^n}$
Solution:
1. Let us put $\small{u = f(ax+b)}$
• Then $\small{\frac{du}{dx}~=~a\,f'(ax+b) \Rightarrow a\,f'(ax+b)\,dx~=~du}$
2. So we want:
$\small{I = \int{\left[f'(ax+b) \left[f(ax+b) \right]^n\right]dx}= \int{\left[\frac{(a)f'(ax+b) \left[f(ax+b) \right]^n}{a}\right]dx}}$
$\small{= \int{\left[\frac{u^n}{a}\right]du}}$
3. This integration gives: $\small{I = \frac{u^{n+1}}{a(n+1)}~+~\rm{C}}$
4. Substituting for u, we get:
$\small{I = \frac{[f(ax+b]^{n+1}}{a(n+1)}~+~\rm{C}}$
Solved Example 23.150
Integrate $\small{\left(\frac{2 + \sin(2x)}{1 + \cos(2x)} \right)e^x}$
Solution:
1. 1. First we will rearrange the given expression:
$\small{\left(\frac{2 + \sin(2x)}{1 + \cos(2x)} \right)e^x~=~\left(\frac{2 + 2\sin(x) \cos (x)}{2 \cos^2 x} \right)e^x}$
$\small{~=~\left(\frac{1}{\cos^2 x}~+~\tan x \right)e^x~=~\left(\sec^2 x~+~\tan x \right)e^x}$
2. So we want:
$\small{I = \int{\left[\left(\sec^2 x~+~\tan x \right)e^x\right]dx}}$
• $\small{\sec^2 x}$ is the derivative of $\small{\tan x}$. So this is of the form:
$\small{\int{\left[\left(f(x)~+~f'(x) \right)e^x\right]dx}~=~e^x\,f(x)~+~\rm{C}}$
3. Thus we get:
$\small{I = \int{\left[\left(\frac{2 + \sin(2x)}{1 + \cos(2x)} \right)e^x\right]dx}= \int{\left[\left(\sec^2 x~+~\tan x \right)e^x\right]dx}~=~e^x\,\tan(x)~+~\rm{C}}$
Solved Example 23.151
Integrate $\small{\frac{\sqrt{x^2 + 1}\left[\log(x^2 + 1)~-~2 \log x \right]}{x^4}}$
Solution:
1. First we will rearrange the given expression:
Put $\small{x~=~\tan u}$
$\small{\Rightarrow 1+x^2~=~1+\tan^2 u~=~\sec^2 u}$
• Also, $\small{\frac{dx}{du}~=~\sec^2 u \Rightarrow dx~=~\sec^2 u \, du}$
• So we can write:
$\small{I~=~\int{\left[\frac{\sqrt{x^2 + 1}\left[\log(x^2 + 1)~-~2 \log x \right]}{x^4} \right]dx}~=~\int{\left[\frac{\sqrt{\sec^2 u}\left[\log(\sec^2 u)~-~ \log (\tan^2 u) \right]}{\tan^4 u} \right]\sec^2 u\,du}}$
$\small{~=~\int{\left[\frac{\left[\log\left(\frac{\sec^2 u}{\tan^2 u} \right) \right]}{\tan^4 u} \right]\sec^3 u\,du}~=~\int{\left[\frac{\left[\log\left(\frac{1}{\sin^2 u} \right) \right]}{\tan^4 u\left(\cos^3 u \right)} \right]du}}$
$\small{~=~\int{\left[\frac{\cos u\left[\log\left(\frac{1}{\sin^2 u} \right) \right]}{\sin^4 u} \right]du}}$
2. We will use the method of integration by parts.
(a) Assigning first and second functions:
♦ Let first function be: $\small{f(u)=\log\left(\frac{1}{\sin^2 u} \right)}$
♦ Let second function be: $\small{g(u)=\frac{\cos u}{\sin^4 u}}$
(b) Finding A:
$\small{A~=~\int{\left[g(u) \right]dx}~=~\frac{- 1}{3\sin^3 u}}$
(c) $\small{\big[f(u) \left(A \right) \big]~=~\big[ \log\left(\frac{1}{\sin^2 u} \right)\big] \,\big[\frac{- 1}{3 \sin^3 u} \big]}$
• This is the first term.
(d) $\small{f'(u)~=~\frac{-2 \cos(u)}{\sin(u)}}$
(e) $\small{\int{\big[f'(u)\,\left(A \right) \big]du}~=~\int{\big[\frac{-2 \cos(u)}{\sin(u)}\,\left(\frac{- 1}{3 \sin^3 (u)} \right)\big]du}}$
$\small{~=~\frac{2}{3}\int{\big[\frac{\cos(u)}{\sin^4(u)}\big]du}~=~\frac{-2}{9\sin^3 u}}$
• This is the second term.
(f) So we get:
$\small{I~=~\text{First term - Second term}}$
$\small{~=~\big[ \log\left(\frac{1}{\sin^2 u} \right)\big] \,\big[\frac{- 1}{3 \sin^3 u} \big]~-~\big[\frac{-2}{9 \sin^3 u} \big]}$
$\small{~=~\big[ \log\left(\frac{1}{\sin^2 u} \right)\big] \,\big[\frac{- 1}{3 \sin^3 u} \big]~+~\big[\frac{2}{9 \sin^3 u} \big]}$
$\small{~=~\big[\frac{2}{9 \sin^3 u} \big]~-~\big[ \log\left(\frac{1}{\sin^2 u} \right)\big] \,\big[\frac{1}{3 \sin^3 u} \big]}$
$\small{~=~\big[\frac{1}{3\sin^3 u} \big]\bigg[\frac{2}{3}~-~ \log\left(\frac{1}{\sin^2 u} \right)\bigg]}$
3. Now we will substitute for u:
(a) We wrote $\small{x~=~\tan u}$
$\small{\Rightarrow 1+x^2~=~1+\tan^2 u~=~\sec^2 u}$
$\small{\Rightarrow \cos^2 u~=~\frac{1}{1+x^2}}$
$\small{\Rightarrow \sin^2 u~=~1~-~\frac{1}{1+x^2}~=~\frac{x^2}{1+x^2}}$
$\small{\Rightarrow \frac{1}{\sin^2 u}~=~1~+~\frac{1}{x^2}}$
$\small{\Rightarrow \frac{1}{\sin u}~=~\left(1~+~\frac{1}{x^2} \right)^{1/2}}$
$\small{\Rightarrow \frac{1}{\sin^3 u}~=~\left(1~+~\frac{1}{x^2} \right)^{3/2}}$
(b) So we get:
$\small{I~=~\big[\frac{1}{3\sin^3 u} \big]\bigg[\frac{2}{3}~-~ \log\left(\frac{1}{\sin^2 u} \right)\bigg]}$
$\small{~=~\big[\frac{1}{3} \left(1~+~\frac{1}{x^2} \right)^{3/2} \big]\bigg[\frac{2}{3}~-~ \log\left(1~+~\frac{1}{x^2} \right)\bigg]~+~\rm{C}}$
Solved Example 23.152
Integrate $\small{\frac{1}{e^x~+~e^{-x}}}$
Solution:
1. First we will rearrange the given expression:
$\small{\frac{1}{e^x~+~e^{-x}}~=~\frac{1}{e^x~+~\frac{1}{e^x}}~=~\frac{e^x}{e^{2x}~+~1}}$
Put $\small{u~=~e^x}$
• Then, $\small{\frac{du}{dx}~=~e^x \Rightarrow e^x\,dx ~=~du}$
2. So we can write:
$\small{I~=~\int{\left[\frac{1}{e^x~+~e^{-x}} \right]dx}~=~\int{\left[\frac{e^x}{e^{2x}~+~1} \right]du}~=~\int{\left[\frac{1}{u^{2}~+~1} \right]du}}$
3. This is a standard integral. We get:
$\small{I~=~\tan^{-1}u~+~\rm{C}}$
4. Substituting for u, we get:
$\small{I~=~\tan^{-1}(e^x)~+~\rm{C}}$
Solved Example 23.153
Evaluate $\small{\int_0^1{\left[e^{2 - 3x} \right]dx}}$ as a limit of a sum.
Solution:
1. 1. $\small{\int_{0}^{1}{\left[e^{2-3x} \right]dx}}$ is the area bounded by the four items:
♦ The curve $\small{y = f(x) = e^{2-3x}}$
♦ The vertical line x = 0 (y-axis)
♦ The vertical line x = 1
♦ The horizontal line y = 0 (x-axis)
2. We have:
$\small{\int_a^b{\left[f(x) \right]dx}~=~\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~.~.~.~+~f(a+nh) \right]}$
• In our present case, a = 0 and b = 1
So $\small{h~=~\frac{b-a}{n}~=~\frac{1-0}{n}~=~\frac{1}{n}}$
3. Now the formula becomes:
Area =
$\small{\lim_{n\rightarrow \infty} \frac{b-a}{n} \left[f(a+h)~+~f(a+2h)~+~f(a+3h)~+~~.~.~.~+f(a+nh) \right]}$
$\small{~=~\lim_{n\rightarrow \infty} \frac{1}{n} \left[f(a+\frac{1}{n})~+~f(a+\frac{2}{n})~+~f(a+\frac{3}{n})~+~~.~.~.~+f(a+(n)\frac{1}{n}) \right]}$
$\small{~=~\lim_{n\rightarrow \infty} \frac{1}{n} \bigg[\big[e^{2-3(a+\frac{1}{n})}\big]~+~\big[e^{2-3(a+\frac{2}{n})}\big]~+~\big[e^{2-3(a+\frac{3}{n})}\big]~+~~.~.~.~}$
$\small{~~~~~~~~~.~.~.~+~\big[e^{2-3(a+(n)\frac{1}{n})}\big]\bigg]}$
4. Let us determine the quantity inside the large square brackets. We have to do a summation:
$\small{e^{2-3(a+\frac{1}{n})}~+~e^{2-3(a+\frac{2}{n})}~+~e^{2-3(a+\frac{3}{n})}~+~.~.~.~ \text{n terms}}$
$\small{~=~\left(e^2~\times~e^{-3a}~\times~e^{\frac{-3}{n}} \right)~+~\left(e^2~\times~e^{-3a}~\times~e^{\frac{-6}{n}} \right)~+~\left(e^2~\times~e^{-3a}~\times~e^{\frac{-9}{n}} \right)~+~.~.~.~ \text{n terms}}$
$\small{~=~e^{2-3a}\left(e^{\frac{-3}{n}}~+~e^{\frac{-6}{n}}~+~e^{\frac{-9}{n}}~+~.~.~.~ \text{n terms} \right)}$
$\small{~=~e^{2}\left(e^{\frac{-3}{n}}~+~e^{\frac{-6}{n}}~+~e^{\frac{-9}{n}}~+~.~.~.~ \text{n terms} \right)}$
$\small{~~~~~~\because {3a}~=~{3(0)}~=~0}$
• So inside the brackets, we have a geometric progression.
• First term = $\small{e^{\frac{-3}{n}}}$
• Common ratio $\small{~r~=~\frac{e^{\frac{-6}{n}}}{e^{\frac{-3}{n}}}~=~\frac{e^{\frac{-9}{n}}}{e^{\frac{-6}{n}}}~=~e^{\frac{-3}{n}}}$
• Sum of the first n terms is given by: $\small{\frac{\text{First term}~\times~\left(r^{n-1}~-~1 \right)}{r~-~1}}$
$\small{r^{n-1}~=~\left(e^{\frac{-3}{n}}\right)^{n-1}~=~e^{\frac{-3n+3}{n}}~=~e^{-3+\frac{3}{n}}}$
• So we get:
Sum of all terms of the G.P
$\small{~=~\frac{e^{\frac{-3}{n}}~\times~\left({e^{-3+\frac{3}{n}}}~-~1 \right)}{e^{\frac{-3}{n}}~-~1}~=~\frac{{e^{-3}}~-~e^{\frac{-3}{n}} }{e^{\frac{-3}{n}}~-~1}}$
• So the summation is:
$\small{e^2 \left[\frac{{e^{-3}}~-~e^{\frac{-3}{n}} }{e^{\frac{-3}{n}}~-~1} \right]}$
5. So the limit in (3) becomes:
$\small{\lim_{n\rightarrow \infty} \frac{1}{n} \Bigg[e^2 \left[\frac{{e^{-3}}~-~e^{\frac{-3}{n}} }{e^{\frac{-3}{n}}~-~1} \right] \Bigg]}$
$\small{~=~e^2 \lim_{n\rightarrow \infty} \frac{1}{n} \left[\frac{{e^{-3}}~-~e^{\frac{-3}{n}} }{e^{\frac{-3}{n}}~-~1} \right]}$
$\small{~=~e^2 \lim_{n\rightarrow \infty} \frac{1}{n} \left[\frac{{e^{-3}} }{e^{\frac{-3}{n}}~-~1} \right]~-~e^2 \lim_{n\rightarrow \infty} \frac{1}{n} \left[\frac{e^{\frac{-3}{n}} }{e^{\frac{-3}{n}}~-~1} \right]}$
$\small{~=~e^2 \lim_{n\rightarrow \infty} \frac{(-3)}{(-3)n} \left[\frac{{e^{-3}} }{e^{\frac{-3}{n}}~-~1} \right]~-~e^2 \lim_{n\rightarrow \infty} \frac{(-3)}{(-3)n} \left[\frac{e^{\frac{-3}{n}} }{e^{\frac{-3}{n}}~-~1} \right]}$
$\small{~=~\frac{e^2}{(-3)} \lim_{n\rightarrow \infty} \frac{-3}{n} \left[\frac{{e^{-3}} }{e^{\frac{-3}{n}}~-~1} \right]~-~\frac{e^2}{(-3)} \lim_{n\rightarrow \infty} \frac{-3}{n} \left[\frac{e^{\frac{-3}{n}} }{e^{\frac{-3}{n}}~-~1} \right]}$
$\small{~=~\frac{e^{-1}}{(-3)} \lim_{n\rightarrow \infty} \frac{-3}{n} \left[\frac{{1} }{e^{\frac{-3}{n}}~-~1} \right]~-~\frac{e^2}{(-3)} \lim_{n\rightarrow \infty} \frac{-3}{n} \left[\frac{e^{\frac{-3}{n}} }{e^{\frac{-3}{n}}~-~1} \right]}$
6. In the above expression, the limits can be evaluated as follows:
$\small{\frac{e^{-1}}{(-3)} \lim_{n\rightarrow \infty} \left[\frac{{1} }{\frac{e^{\frac{-3}{n}}~-~1}{\frac{-3}{n}}} \right]
~-~\frac{e^2}{(-3)} \lim_{n\rightarrow \infty} \left[\frac{{e^{\frac{-3}{n}}} }{\frac{e^{\frac{-3}{n}}~-~1}{\frac{-3}{n}}} \right]}$
$\small{~=~\frac{e^{-1}}{(-3)} \bigg[\frac{1}{1} \bigg]~-~\frac{e^2}{(-3)}\big[\frac{1}{1} \big]~=~\frac{e^{-1}~-~e^2}{(-3)}}$
$\small{~=~\frac{e^{2}~-~e^{-1}}{3}}$
• Here we use two facts:
(i) $\small{\lim_{n\rightarrow \infty} \left[e^{\frac{-3}{n}} \right]~=~e^{\frac{-3}{\infty}}~=~e^0~=~1}$
(ii) Let $\small{\frac{-3}{n}~=~h}$.
Then $\small{h \rightarrow 0 ~\text{as}~n \rightarrow \infty}$
So $\small{\lim_{n\rightarrow \infty} \Big[\frac{e^{\frac{-3}{n}}~-~1}{\frac{-3}{n}}\Big]~=~\lim_{n\rightarrow \infty} \Big[\frac{e^{h}~-~1}{h}\Big]~=~1}$
Solved Example 23.154
Evaluate $\small{\int_{0}^{\frac{\pi}{4}}{\left[\frac{\sin x \cos x}{\cos^4 x + \sin^4 x} \right]dx}}$
Solution:
1. Let us rearrange the given expression:
$\small{\frac{\sin x \cos x}{\cos^4 x + \sin^4 x}~=~\frac{\sin x \cos x}{\cos^4 x + \sin^4 x~+~2\sin^2x\,\cos^2x~-~2\sin^2x\,\cos^2x}}$
$\small{~=~\frac{\sin x \cos x}{\left(\cos^2 x + \sin^2 x \right)^2~-~2\sin^2x\,\cos^2x}~=~\frac{\sin x \cos x}{1~-~2\sin^2x\,\cos^2x}}$
$\small{~=~\frac{\sin x \cos x}{1~-~(4\sin^2x\,\cos^2x)/2}~=~\frac{2\sin x \cos x}{2~-~4\sin^2x\,\cos^2x}}$
$\small{~=~\frac{\sin (2x)}{2~-~\sin^2(2x)}~=~\frac{\sin (2x)}{1+[1~-~\sin^2(2x)]}~=~\frac{\sin (2x)}{1+[\cos^2(2x)]}}$
2. So we want: $\small{I~=~\int_{0}^{\frac{\pi}{4}}{\left[\frac{\sin x \cos x}{\cos^4 x + \sin^4 x} \right]dx}~=~\int_{0}^{\frac{\pi}{4}}{\left[\frac{\sin (2x)}{1+\cos^2(2x)} \right]dx}}$
3. First we will find the indefinite integral:
$\small{F~=~\int{\left[\frac{\sin (2x)}{1+\cos^2(2x)} \right]dx}}$
• Put $\small{u = \cos(2x)}$
Then $\small{\frac{du}{dx}~=~-2\sin(2x) \Rightarrow -2\sin(2x)\,dx~=~du}$
• We want:
$\small{F~=~\int{\left[\frac{(-2)\sin (2x)}{(-2)\left[1+\cos^2(2x) \right]} \right]dx}~=~\int{\left[\frac{1}{(-2)\left[1+u^2 \right]} \right]dx}~=~\frac{1}{(-2)}\int{\left[\frac{1}{1+u^2 } \right]dx}}$
• This is a standard integral. We get:
$\small{F~=~\left(\frac{-1}{2} \right)\tan^{-1}u}$
• Subsituting for u, we get:
$\small{F~=~\left(\frac{-1}{2} \right)\tan^{-1}\left[\cos(2x) \right]}$
4. Now we can evaluate the definite integral. We get:
$\small{I~=~\int_{0}^{\frac{\pi}{4}}{\left[\frac{\sin (2x)}{1+\cos^2(2x)} \right]dx}}$
$\small{~=~\big[\left(\frac{-1}{2} \right)\tan^{-1}\left[\cos(2x) \right]\big]_{0}^{\frac{\pi}{4}}}$
$\small{~=~\left(\frac{-1}{2} \right) \big[\tan^{-1}\left[\cos(2x) \right]\big]_{0}^{\frac{\pi}{4}}}$
$\small{~=~\left(\frac{-1}{2} \right) \big[\tan^{-1}\left[\cos\left(\frac{\pi}{2} \right) \right]~-~\tan^{-1}\left[\cos\left(0 \right) \right]\big]}$
$\small{~=~\left(\frac{-1}{2} \right) \big[\tan^{-1}\left[0 \right]~-~\tan^{-1}\left[1 \right]\big]}$
$\small{~=~\left(\frac{-1}{2} \right) \big[0~-~\frac{\pi}{4}\big]~=~\frac{\pi}{8}}$
Solved Example 23.155
Evaluate $\small{\int_{\frac{\pi}{2}}^{\pi}{\left[e^x\left(\frac{1-\sin x}{1 - \cos x} \right) \right]dx}}$
Solution:
1. Let us rearrange the portion inside braces:
$\small{\frac{1-\sin x}{1 - \cos x}~=~\frac{1-\sin x}{2 \sin^2\left(\frac{x}{2} \right)}~=~\frac{1}{2 \sin^2\left(\frac{x}{2} \right)}~-~\frac{\sin x}{2 \sin^2\left(\frac{x}{2} \right)}}$
$\small{~=~\frac{1}{2 \sin^2\left(\frac{x}{2} \right)}~-~\frac{2 \sin\left(\frac{x}{2} \right) \cos\left(\frac{x}{2} \right)}{2 \sin^2\left(\frac{x}{2} \right)}}$
$\small{~=~\frac{\csc^2\left(\frac{x}{2} \right)}{2}~-~\cot\left(\frac{x}{2} \right)~=~(-1)\left[\cot\left(\frac{x}{2} \right)~-~\frac{\csc^2\left(\frac{x}{2} \right)}{2} \right]}$
$\small{~=~(-1)\left[\cot\left(\frac{x}{2} \right)~+~\frac{(-1)\csc^2\left(\frac{x}{2} \right)}{2} \right]}$
2. So the given expression can be written as:
$\small{(-1)\,e^x\left[\cot\left(\frac{x}{2} \right)~+~\frac{(-1)\csc^2\left(\frac{x}{2} \right)}{2} \right]}$
• In the above result, $\small{\left[\frac{(-1)\csc^2\left(\frac{x}{2} \right)}{2} \right]}$ is the derivative of $\small{\left[\cot\left(\frac{x}{2} \right) \right]}$
• So the given expression can be written as:
$\small{(-1)\,e^x\left[f(x)~+~f'(x) \right]}$
3. So the indefinite integral can be written as:
$\small{F~=~(-1)\,e^x\left[f(x)\right]~=~(-1)\,e^x\left[\cot\left(\frac{x}{2} \right)\right]}$
4. Now we can evaluate the definite integral. We get:
$\small{I~=~\int_{\frac{\pi}{2}}^{\pi}{\left[e^x\left(\frac{1-\sin x}{1 - \cos x} \right) \right]dx}}$
$\small{~=~\big[(-1)\,e^x\left[\cot\left(\frac{x}{2} \right)\right]\big]_{\frac{\pi}{2}}^{\pi}}$
$\small{~=~\big[(-1)\,e^{\pi}\left[\cot\left(\frac{\pi}{2} \right)\right]\big]~-~\big[(-1)\,e^{\frac{\pi}{2}}\left[\cot\left(\frac{\pi}{4} \right)\right]\big]}$
$\small{~=~\big[(-1)\,e^{\pi}\left[0\right]\big]~-~\big[(-1)\,e^{\frac{\pi}{2}}\left[1\right]\big]}$
$\small{~=~e^{\frac{\pi}{2}}}$
Solved Example 23.156
Evaluate $\small{\int_{0}^{\frac{\pi}{2}}{\left[\frac{\cos^2 x}{\cos^2 x ~+~ 4 \sin^2 x} \right]dx}}$
Solution:
1. Let us rearrange the given expression:
$\small{\frac{\cos^2 x}{\cos^2 x ~+~ 4 \sin^2 x}~=~\frac{\cos^2 x}{\cos^2 x ~+~ 4(1-\cos^2 x)}~=~\frac{\cos^2 x}{4~-~3 \cos^2 x}}$
$\small{~=~\left(\frac{-1}{3} \right)\left[\frac{(-3)\cos^2 x}{4~-~3 \cos^2 x} \right]~=~\left(\frac{-1}{3} \right)\left[\frac{4~-~3\cos^2 x~-~4}{4~-~3 \cos^2 x} \right]}$
$\small{~=~\left(\frac{-1}{3} \right)\left[1~-~\frac{4}{4~-~3 \cos^2 x} \right]~=~\left(\frac{-1}{3} \right)\left[1~-~\frac{\frac{4}{\cos^2 x}}{\frac{4}{\cos^2 x}~-~\frac{3\cos^2 x}{\cos^2 x}} \right]}$
$\small{~=~\left(\frac{-1}{3} \right)\left[1~-~\frac{4 \sec^2 x}{4 \sec^2 x~-~3} \right]~=~\left(\frac{-1}{3} \right)\left[1~-~\frac{4 \sec^2 x}{4 (1+\tan^2 x)~-~3} \right]}$
$\small{~=~\left(\frac{-1}{3} \right)\left[1~-~\frac{4 \sec^2 x}{1~+~4 \tan^2 x} \right]}$
2. Now we can find the indefinite integral:
$\small{F~=~\int{\left[\frac{\cos^2 x}{\cos^2 x ~+~ 4 \sin^2 x} \right]dx}~=~\int{\bigg[\left(\frac{-1}{3} \right)\left[1~-~\frac{4 \sec^2 x}{1~+~4 \tan^2 x} \right] \bigg]dx}}$
$\small{~=~\int{\bigg[\left(\frac{-1}{3} \right) \bigg]dx}~+~\left(\frac{1}{3} \right)\int{\bigg[\frac{4 \sec^2 x}{1~+~4 \tan^2 x} \bigg]dx}}$
$\small{~=~\frac{-x}{3}~+~\left(\frac{1}{3} \right)\int{\bigg[\frac{4 \sec^2 x}{1~+~4 \tan^2 x} \bigg]dx}}$
$\small{~=~F_1~+~F_2}$
• $\small{F_2}$ can be calculated as follows:
• Put $\small{u~=~\tan x}$. Then $\small{\frac{du}{dx}~=~\sec^2 x \Rightarrow \sec^2 x\,dx~=~du}$
• So we get:
$\small{F_2~=~\left(\frac{1}{3} \right)\int{\bigg[\frac{4}{1~+~4(u)^2} \bigg]dx}~=~\left(\frac{4}{3} \right)\int{\bigg[\frac{1}{1~+~(2u)^2} \bigg]dx}}$
$\small{~=~\left(\frac{4}{3} \right)\left(\frac{1}{2} \right)\bigg[\tan^{-1}(2u) \bigg]~=~\left(\frac{2}{3} \right)\bigg[\tan^{-1}(2 \tan x) \bigg]}$
• Therefore, $\small{F~=~\frac{-x}{3}~+~\left(\frac{2}{3} \right)\left[\tan^{-1}(2 \tan x) \right]}$
3. Now we can evaluate the definite integral. We get:
$\small{I~=~\big[\frac{-x}{3}~+~\left(\frac{2}{3} \right)\left[\tan^{-1}(2 \tan x) \right]\big]_{0}^{\frac{\pi}{2}}}$
$\small{~=~\big[\frac{-\pi}{6}~+~\left(\frac{2}{3} \right)\left[\tan^{-1}(2 \tan \frac{\pi}{2}) \right]\big]~-~\big[0~+~\left(\frac{2}{3} \right)\left[\tan^{-1}(2 \tan 0) \right]\big]}$
$\small{~=~\big[\frac{-\pi}{6}~+~\left(\frac{2}{3} \right)\left[\tan^{-1}(\infty) \right]\big]~-~\big[0~+~\left(\frac{2}{3} \right)\left[\tan^{-1}(0) \right]\big]}$
$\small{~=~\big[\frac{-\pi}{6}~+~\left(\frac{2}{3} \right)\left[\frac{\pi}{2} \right]\big]~-~\big[0~+~\left(\frac{2}{3} \right)\left[0 \right]\big]}$
$\small{~=~\big[\frac{-\pi}{6}~+~\frac{2\pi}{6}\big]~=~\frac{\pi}{6}}$
Solved Example 23.157
Evaluate $\small{\int_{0}^{1}{\left[\frac{1}{\sqrt{1+x}~-~\sqrt{x}} \right]dx}}$
Solution:
1. Let us rearrange the given expression:
$\small{\frac{1}{\sqrt{1+x}~-~\sqrt{x}}~=~\frac{1(\sqrt{1+x}~+~\sqrt{x})}{(\sqrt{1+x}~-~\sqrt{x})(\sqrt{1+x}~+~\sqrt{x})}}$
$\small{~=~\frac{\sqrt{1+x}~+~\sqrt{x}}{1+x~-~x}~=~\sqrt{1+x}~+~\sqrt{x}}$
2. So we can write:
$\small{\int_{0}^{1}{\left[\frac{1}{\sqrt{1+x}~-~\sqrt{x}} \right]dx}~=~\int_{0}^{1}{\left[\sqrt{1+x}~+~\sqrt{x} \right]dx}}$
3. Now we can write the indefinite integral:
$\small{F~=~\int{\left[\sqrt{1+x}~+~\sqrt{x} \right]dx}~=~\frac{(1+x)^{3/2}}{3/2}~+~\frac{(x)^{3/2}}{3/2}}$
$\small{~=~\frac{(1+x)^{3/2}~+~(x)^{3/2}}{3/2}}$
$\small{I~=~\int_{0}^{1}{\left[\sqrt{1+x}~+~\sqrt{x} \right]dx}}$
$\small{~=~\big[\frac{(1+x)^{3/2}~+~(x)^{3/2}}{3/2}\big]_{0}^{1}}$
$\small{~=~\big[\frac{(1+1)^{3/2}~+~(1)^{3/2}}{3/2}\big]~-~\big[\frac{(1+0)^{3/2}~+~(0)^{3/2}}{3/2}\big]}$
$\small{~=~\big[\frac{2^{3/2}~+~1}{3/2}\big]~-~\big[\frac{1~+~0}{3/2}\big]~=~\frac{2^{3/2}}{3/2}~=~\frac{2^{5/2}}{3}~=~\frac{4 \sqrt2}{3}}$
Solved Example 23.158
Evaluate $\small{\int_{0}^{\pi}{\left[\frac{x \tan x}{\sec x~+~\tan x} \right]dx}}$
Solution:
1. Let us rearrange the given expression:
$\small{\frac{x \tan x}{\sec x~+~\tan x}~=~\frac{x \frac{\sin x}{\cos x}}{\frac{1}{\cos x}~+~\frac{\sin x}{\cos x}}~=~\frac{x \sin x}{1~+~\sin x}}$
2. So we want:
$\small{I~=~\int_{0}^{\pi}{\left[\frac{x \tan x}{\sec x~+~\tan x} \right]dx}~=~\int_{0}^{\pi}{\left[\frac{x \sin x}{1~+~\sin x} \right]dx}}$
3. Applying P4, we get:
$\small{I~=~\int_{0}^{\pi}{\left[\frac{(\pi-x) \sin (\pi-x)}{1~+~\sin (\pi-x)} \right]dx}~=~\int_{0}^{\pi}{\left[\frac{(\pi-x) \sin x}{1~+~\sin x} \right]dx}}$
$\small{~=~\int_{0}^{\pi}{\left[\frac{\pi \sin x}{1~+~\sin x} \right]dx}~-~\int_{0}^{\pi}{\left[\frac{ x \sin x}{1~+~\sin x} \right]dx}}$
$\small{\Rightarrow I~=~\int_{0}^{\pi}{\left[\frac{\pi \sin x}{1~+~\sin x} \right]dx}~-~I}$
$\small{\Rightarrow 2I~=~\int_{0}^{\pi}{\left[\frac{\pi \sin x}{1~+~\sin x} \right]dx}}$
$\small{\Rightarrow I~=~\frac{\pi}{2}\int_{0}^{\pi}{\left[\frac{ \sin x}{1~+~\sin x} \right]dx}}$
$\small{\Rightarrow I~=~\frac{\pi}{2}\int_{0}^{\pi}{\left[\frac{\sin x (1~-~\sin x)}{(1~+~\sin x)(1~-~\sin x)} \right]dx}}$
$\small{~=~\frac{\pi}{2} \int_{0}^{\pi}{\left[\frac{\sin x~-~\sin^2 x}{1~-~\sin^2 x} \right]dx}~=~\frac{\pi}{2} \int_{0}^{\pi}{\left[\frac{\sin x~-~\sin^2 x}{\cos^2 x} \right]dx}}$
$\small{~=~\frac{\pi}{2} \int_{0}^{\pi}{\left[\sec x\, \tan x~-~\tan^2 x \right]dx}}$
$\small{~=~\frac{\pi}{2} \int_{0}^{\pi}{\left[\sec x\, \tan x~-~\left(\sec^2 x - 1 \right) \right]dx}}$
$\small{\Rightarrow I~=~\frac{\pi}{2} \int_{0}^{\pi}{\left[\sec x\, \tan x \right]dx}~-~\frac{\pi}{2} \int_{0}^{\pi}{\left[\sec^2 x \right]dx}~+~\frac{\pi}{2} \int_{0}^{\pi}{\left[1 \right]dx}}$
$\small{\Rightarrow I~=~I_1~-~I_2~+~I_3}$
4. First we will calculate I1:
$\small{F_1~=~\frac{\pi}{2} \int{\left[\sec x\, \tan x \right]dx}~=~\frac{\pi}{2}\,\sec x}$
$\small{\Rightarrow I_1~=~\left[\frac{\pi}{2}\,\sec x \right]_{0}^{\pi}~=~\left[\frac{\pi}{2}\,\sec \left(\pi \right) \right]~-~\left[\frac{\pi}{2}\,\sec \left(0 \right) \right]}$
$\small{\Rightarrow I_1~=~\left[\frac{-\pi}{2} \right]~-~\left[\frac{\pi}{2} \right]~=~-\pi}$
5. Next we will calculate I2:
$\small{F_2~=~\frac{\pi}{2} \int{\left[\sec^2 x \right]dx}~=~\frac{\pi}{2}\,\tan x}$
$\small{\Rightarrow I_2~=~\left[\frac{\pi}{2}\,\tan x \right]_{0}^{\pi}~=~\left[\frac{\pi}{2}\,\tan \left(\pi \right) \right]~-~\left[\frac{\pi}{2}\,\tan \left(0 \right) \right]}$
$\small{\Rightarrow I_2~=~\left[0 \right]~-~\left[0 \right]~=~0}$
6. Finally we will calculate I3:
$\small{F_3~=~\frac{\pi}{2} \int{\left[1\right]dx}~=~\frac{\pi}{2}\,x}$
$\small{\Rightarrow I_3~=~\left[\frac{\pi}{2}\,(x) \right]_{0}^{\pi}~=~\left[\frac{\pi}{2}\, \left(\pi \right) \right]~-~\left[\frac{\pi}{2} \left(0 \right) \right]~=~\frac{\pi^2}{2}}$
7. From (3), (4), (5) and (6), we get:
$\small{I~=~I_1~-~I_2~+~I_3}$
$\small{~=~-\pi~-~0~+~\frac{\pi^2}{2}}$
$\small{~=~\left(\frac{\pi^2}{2}~-~\pi \right)~=~\left(\frac{\pi^2}{2}~-~\frac{2 \pi}{2} \right)}$
$\small{~=~\frac{\pi}{2}\left(\pi~-~2 \right)}$
Solved Example 23.159
Evaluate $\small{\int_{0}^{\frac{\pi}{4}}{\left[\frac{\sin x~+~\cos x}{9~+~16 \sin(2x)} \right]dx}}$
Solution:
1. Put $\small{u = \sin x ~-~\cos x}$
• Then $\small{\frac{du}{dx}~=~\cos x ~+~\sin x}$
$\small{\Rightarrow dx(\cos x ~+~\sin x)~=~du}$
• Also, $\small{u^2~=~\sin^2 x~+~\cos^2 x~-~2 \sin x \cos x~=~1~-~2 \sin x \cos x}$
$\small{\Rightarrow u^2~=~1~-~\sin(2x)}$
$\small{\Rightarrow \sin(2x)~=~1~-~u^2}$
2. So we want:
$\small{F ~=~\int{\left[\frac{\sin x~+~\cos x}{9~+~16 \sin(2x)} \right]dx}~=~\int{\left[\frac{1}{9~+~16(1-u^2)} \right]du}}$
$\small{~=~\int{\left[\frac{1}{9~+~16-16u^2} \right]du}~=~\int{\left[\frac{1}{25-16u^2} \right]du}~=~\int{\left[\frac{1}{5^2-(4u)^2} \right]du}}$
3. This is a standard integral. We have:
$\small{\int{\left[\frac{1}{a^2~-~t^2} \right]dt}~=~\frac{1}{2a} \log \left| \frac{a+t}{a-t} \right|~+~\rm{C}}$
In our present case, a = 5 and t = 4u
• So we get:
$\small{F~=~\int{\left[\frac{1}{5^2-(4u)^2} \right]du}~=~\frac{1}{2(5)(4)} \log \left| \frac{5+4u}{5-4u} \right|~=~\frac{1}{40} \log \left| \frac{5+4u}{5-4u} \right|}$
4. Substituting for u, we get:
$\small{F~=~\frac{1}{40} \log \left|\frac{5~+~4 \sin x~-~4 \cos x}{5~-~4 \sin x~+~4 \cos x} \right|}$
5. Now we can find the definite integral:
$\small{I~=~\left[\frac{1}{40} \log \left|\frac{5~+~4 \sin x~-~4 \cos x}{5~-~4 \sin x~+~4 \cos x} \right| \right]_0^{\frac{\pi}{4}}}$
$\small{~=~\left[\frac{1}{40} \log \left|\frac{5~+~4 \sin \left(\frac{\pi}{4} \right)~-~4 \cos \left(\frac{\pi}{4} \right)}{5~-~4 \sin \left(\frac{\pi}{4} \right)~+~4 \cos \left(\frac{\pi}{4} \right)} \right| \right]~-~\left[\frac{1}{40} \log \left|\frac{5~+~4 \sin \left(0 \right)~-~4 \cos \left(0 \right)}{5~-~4 \sin \left(0 \right)~+~4 \cos \left(0 \right)} \right| \right]}$
$\small{~=~\left[\frac{1}{40} \log \left|\frac{5~+~0}{5~-~0} \right| \right]~-~\left[\frac{1}{40} \log \left|\frac{5~+~0~-~4}{5~-~0~+~4} \right| \right]}$
$\small{~=~\left[\frac{1}{40} \log \left|1 \right| \right]~-~\left[\frac{1}{40} \log \left|\frac{1}{9} \right| \right]}$
$\small{~=~\left[\frac{1}{40} (0) \right]~-~\left[\frac{1}{40} \log \left|\frac{1}{9} \right| \right]~=~~-~\left[\frac{1}{40} \log \left|\frac{1}{9} \right| \right]}$
$\small{~=~\frac{1}{40} \log \left(9 \right)}$
Solved Example 23.160
Evaluate $\small{\int_{0}^{\frac{\pi}{2}}{\left[\sin(2x) \tan^{-1}(\sin x) \right]dx}}$
Solution:
1. Put $\small{u = \sin x}$
• Then $\small{\frac{du}{dx}~=~\cos x \Rightarrow dx~=~\frac{du}{\cos x}}$
• We wrote: $\small{u = \sin x}$
♦ When x approaches 0, u approaches 0
♦ When x approaches $\small{\frac{\pi}{2}}$, u approaches 1
2. We want:
$\small{I = \int_{0}^{\frac{\pi}{2}}{\left[\sin(2x) \tan^{-1}(\sin x) \right]dx}}$
$\small{= \int_{0}^{\frac{\pi}{2}}{\left[2 \sin(x)\cos(x) \tan^{-1}(\sin x) \right]dx}}$
$\small{= \int_{0}^{\frac{\pi}{2}}{\left[2 u\cos(x) \tan^{-1}(u) \right]\frac{du}{\cos x}}}$
$\small{= \int_{0}^{1}{\left[2 u \tan^{-1}(u) \right]du}= 2 \int_{0}^{1}{\left[u \tan^{-1}(u) \right]du}}$
3. We will apply the method of integration by parts
(a) Assigning first and second functions:
♦ Let first function be: $\small{f(u)=\tan^{-1}u}$
♦ Let second function be: $\small{g(u)=u}$
(b) Finding A:
$\small{A~=~\int{\left[g(u) \right]dx}~=~\frac{u^2}{2}}$
(c) $\small{\big[f(u) \left(A \right) \big]~=~\big[ \tan^{-1}u\big] \,\big[\frac{u^2}{2} \big]~=~\big[\frac{u^2\,\tan^{-1}u}{2} \big]}$
• This is the first term.
(d) $\small{f'(u)~=~\frac{1}{1 + u^2}}$
(e) $\small{\int{\big[f'(u)\,\left(A \right) \big]du}~=~\int{\big[\frac{1}{1 + u^2}\,\left(\frac{u^2}{2} \right)\big]du}}$
$\small{~=~\int{\big[\frac{u^2}{2(1 + u^2)}\big]du}~=~\frac{1}{2}\int{\big[\frac{u^2}{1 + u^2}\big]du}}$
$\small{~=~\frac{1}{2}\int{\big[\frac{u^2 + 1 - 1}{1 + u^2}\big]du}~=~\frac{1}{2}\int{\big[1~-~\frac{1}{1 + u^2}\big]du}}$
$\small{~=~\frac{1}{2}\int{\big[1\big]du}~-~\frac{1}{2}\int{\big[\frac{1}{1 + u^2}\big]du}}$
$\small{~=~\frac{u}{2}~-~\frac{\tan^{-1}u}{2}}$
• This is the second term.
(f) So we get:
$\small{\int_{0}^{\frac{1}{2}}{\left[u \tan^{-1}(u) \right]du}~=~\text{First term - Second term}}$
$\small{~=~\big[\frac{u^2\,\tan^{-1}u}{2}\big]~-~\big[\frac{u}{2}~-~\frac{\tan^{-1}u}{2} \big]}$
$\small{~=~\frac{u^2\,\tan^{-1}u}{2}~-~\frac{u}{2}~+~\frac{\tan^{-1}u}{2}}$
$\small{~=~\frac{(u^2 + 1)(\tan^{-1}u)~-~u}{2}}$
4. So from (2), we get:
$\small{I= 2 \int_{0}^{1}{\left[u \tan^{-1}(u) \right]du}~=~2 \big[\frac{(u^2 + 1)(\tan^{-1}u)~-~u}{2} \big]_0^1}$
$\small{~=~ \big[(u^2 + 1)(\tan^{-1}u)~-~u \big]_0^1}$
$\small{~=~ \big[(1^2 + 1)(\tan^{-1}1)~-~1 \big]~-~\big[(0 + 1)(\tan^{-1}0)~-~0 \big]}$
$\small{~=~ \big[(2)(\pi/4)~-~1 \big]~-~\big[(1)(0)~-~0 \big]}$
$\small{~=~ \frac{\pi}{2}~-~1}$
The link below gives a few more miscellaneous examples:
Miscellaneous Exercise
In the next chapter, we will see applications of integrals.
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