23.31 - Miscellaneous Examples (3) on Integrals

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

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

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

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

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

2. So we want:

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

3. This is a standard integral of the form:

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

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

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

4. Substituting for u, we get:

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

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

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

2. So we want:

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

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

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

4. Substituting for u, we get:

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

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

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

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

2. So we want:

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

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

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

3. Thus we get:

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

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

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

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

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

• So we can write:

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

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

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

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

(a) Assigning first and second functions:

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

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

(b) Finding A:

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

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

• This is the first term.

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

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

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

• This is the second term.

(f) So we get:

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

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

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

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

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

3. Now we will substitute for u:

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

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

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

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

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

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

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

(b) So we get:

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

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

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

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

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

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

2. So we can write:

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

3. This is a standard integral. We get:

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

4. Substituting for u, we get:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

• So we get:

Sum of all terms of the G.P

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

• So the summation is:

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

5. So the limit in (3) becomes:

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

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

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

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

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

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

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

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

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

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

• Here we use two facts:

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

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

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

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

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

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

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

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

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

3. First we will find the indefinite integral:

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

• Put $\small{u = \cos(2x)}$

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

• We want:

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

• This is a standard integral. We get:

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

• Subsituting for u, we get:

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

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

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

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

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

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

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

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

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

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

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

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

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

2. So the given expression can be written as:

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

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

• So the given expression can be written as:

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

3. So the indefinite integral can be written as:

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

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

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

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

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

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

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

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

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

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

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

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

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

2. Now we can find the indefinite integral:

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


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

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

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

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

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

• So we get:

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

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

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

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

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

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

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

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

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

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

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

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

2. So we can write:

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

3. Now we can write the indefinite integral:

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

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

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

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

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

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

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

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

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

2. So we want:

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

3. Applying P₄, we get:

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

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

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

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

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

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

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

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

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

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

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

4. First we will calculate I₁:

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

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

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

5. Next we will calculate I₂:

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

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

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

6. Finally we will calculate I₃:

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

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

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

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

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

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

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

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

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

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

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

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

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

2. So we want:

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

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

3. This is a standard integral. We have:

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

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

• So we get:

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

4. Substituting for u, we get:

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

5. Now we can find the definite integral:

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

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

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

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

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

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

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

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

• We wrote: $\small{u = \sin x}$

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

2. We want:

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

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

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

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

3. We will apply the method of integration by parts

(a) Assigning first and second functions:

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

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

(b) Finding A:

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

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

• This is the first term.

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

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

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

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

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

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

• This is the second term.

(f) So we get:

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

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

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

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

4. So from (2), we get:

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

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

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

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

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

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The link below gives a few more miscellaneous examples:

Miscellaneous Exercise

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

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23.30 - Miscellaneous Examples (2) on Integrals

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

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

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

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

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

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

3. This integration gives:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

3. This integration gives:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

3. Next we will calculate I₁:

(a) Assigning first and second functions:

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

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

(b) Finding A:

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

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

• This is the first term.

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

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

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

• This is the second term.

(f) So we get:

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

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

4. Next we calculate I₂:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

3. Next we calculate I₁. We get:

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

4. Next we calculate I₂:

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

5. So we can write:

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

• Substituting for u, we get:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

2. So we get:

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

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

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

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

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

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

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

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

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

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

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

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

2. So we get:

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

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

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

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

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

4. So we want:

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

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

5. This integration gives:

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

6. Substituting for u, we get:

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

7. From step (1), we have:

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

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

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

• So from (6), we get:

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

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

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

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

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

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

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

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

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

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

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

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

So we get:

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

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

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

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

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

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

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

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

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

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

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

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

3. This integration gives:

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

4. Substituting for u, we get:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

9. So the integration becomes easy. We get:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

9. So the integration becomes easy. We get:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

9. So the integration becomes easy. We get:

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

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

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

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

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

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

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

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

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

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

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

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

• Let us write:

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

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

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

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

2. So the given expression becomes:

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

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

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

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

5. So we want:

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

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

7. Substituting for t, we get:

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

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

• Let us write:

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

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

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

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

2. So the given expression becomes:

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

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

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

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

5. So we want:

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

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

7. Substituting for t, we get:

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

---

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

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23.20 - Solved Examples on Standard Integrals

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

Solved Example 23.63
Find $\small{\int{\left[\sqrt{4 - x^2} \right]dx}}$
Solution:
1. The given integral can be written as: $\small{\int{\left[\sqrt{2^2 - x^2} \right]dx}}$

2. Now we can use the standard integral:

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

Here a = 2

• So we get:

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

$\small{\Rightarrow \int{\left[\sqrt{4 - x^2} \right]dx}~=~~\frac{x}{2}\sqrt{4 - x^2} ~+~2 \sin^{-1}\frac{x}{2}~+~\rm{C}}$

Solved Example 23.64
Find $\small{\int{\left[\sqrt{1 - 4 x^2} \right]dx}}$
Solution:
1. The given integral can be written as:

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

2. Put u = 2x Then du/dx = 2 ⇒ du = 2 dx

• So the given integral becomes:

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

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

3. Now we can use the standard integral:

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

Here a = 1

• So we get:

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

4. Substituting for u, we get:

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

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

Solved Example 23.65
Find $\small{\int{\left[\sqrt{x^2 + 4x + 1} \right]dx}}$
Solution:
1. The given integral can be rearranged as shown below:

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

2. Put u = x+2 Then du/dx = 1 ⇒ du = dx

So the given integral becomes: $\small{\int{\left[\sqrt{u^2 - (\sqrt{3})^2} \right]du}}$

3. Now we can use the standard integral:

$\bf{\int{\big[\sqrt{x^2-a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 - a^2} ~-~\frac{a^2}{2} \log \left|x+\sqrt{x^2 - a^2} \right|~+~\rm{C}}$

Here x = u and a = √3

• So we get:

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

4. Substituting for u, we get:

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

$\small{\Rightarrow \int{\left[\sqrt{x^2 + 4x + 1} \right]dx}~=~\frac{x+2}{2}\sqrt{x^2 + 4x + 1} ~-~ \frac{3}{2} \log \left|x+2-\sqrt{x^2 + 4x + 1} \right|~+~\rm{C}}$

Solved Example 23.66
Find $\small{\int{\left[\sqrt{x^2 + 4x - 5} \right]dx}}$
Solution:
1. The given integral can be rearranged as shown below:

$\small{\int{\left[\sqrt{x^2 +4x+ 4 - 9} \right]dx}~=~\int{\left[\sqrt{(x+2)^2 - 3^2} \right]dx}}$

2. Put u = x+2 Then du/dx = 1 ⇒ du = dx

So the given integral becomes: $\small{\int{\left[\sqrt{u^2 - 3^2} \right]du}}$

3. Now we can use the standard integral:

$\bf{\int{\big[\sqrt{x^2-a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 - a^2} ~-~\frac{a^2}{2} \log \left|x+\sqrt{x^2 - a^2} \right|~+~\rm{C}}$

Here x = u and a = 3

• So we get:

$\small{\int{\left[\sqrt{u^2 - 3^2} \right]du}~=~\frac{u}{2}\sqrt{u^2 - 3^2} ~-~\frac{3^2}{2} \log \left|u+\sqrt{u^2 - 3^2} \right|~+~\rm{C}}$

4. Substituting for u, we get:

$\small{\int{\left[\sqrt{(x+2)^2 - 3^2} \right]dx}}$

$\small{~=~\frac{x+2}{2}\sqrt{(x+2)^2 - 3^2} ~-~\frac{3^2}{2} \log \left|(x+2)-\sqrt{(x+2)^2 + 3^2} \right|~+~\rm{C}}$

$\small{\Rightarrow \int{\left[\sqrt{x^2 + 4x - 5} \right]dx}}$

$\small{~=~\frac{x+2}{2}\sqrt{x^2 + 4x - 5} ~-~ \frac{9}{2} \log \left|x+2-\sqrt{x^2 + 4x - 5} \right|~+~\rm{C}}$

Solved Example 23.67
Find $\small{\int{\left[\sqrt{3 - 2x - x^2}\right]dx}}$
Solution:
1. The given integral can be rearranged as shown below:

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

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

2. Put u = x+1 Then du/dx = 1 ⇒ du = dx

So the given integral becomes: $\small{\int{\left[\sqrt{2^2 - u^2} \right]du}}$

3. Now we can use the standard integral:

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

Here a = 2 and x = u

• So we get:

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

4. Substituting for u, we get:

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

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

$\small{\Rightarrow \int{\left[\sqrt{3-2x-x^2} \right]dx}}$

$\small{~=~\frac{x+1}{2}\sqrt{3-2x-x^2}~+~2 \sin^{-1}\frac{x+1}{2}~+~\rm{C}}$

Solved Example 23.68
Find $\small{\int{\left[\sqrt{1 - 4x - x^2}\right]dx}}$
Solution:
1. The given integral can be rearranged as shown below:

$\small{\int{\left[\sqrt{-(x^2 +4x- 1)} \right]dx}~=~\int{\left[\sqrt{-[x^2 +4x+4-5]}\right]dx}}$

$\small{~=~\int{\left[\sqrt{-\left[(x+2)^2-(\sqrt{5})^2\right]}\right]dx}~=~\int{\left[\sqrt{(\sqrt{5})^2 - (x+2)^2}\right]dx}}$

2. Put u = x+2 Then du/dx = 1 ⇒ du = dx

So the given integral becomes: $\small{\int{\left[\sqrt{(\sqrt{5})^2 - u^2} \right]du}}$

3. Now we can use the standard integral:

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

Here a = √5 and x = u

• So we get:

$\small{\int{\left[\sqrt{(\sqrt{5})^2 - u^2} \right]dx}~=~~\frac{u}{2}\sqrt{(\sqrt{5})^2 - u^2} ~+~\frac{(\sqrt{5})^2}{2} \sin^{-1}\frac{u}{\sqrt{5}}~+~\rm{C}}$

4. Substituting for u, we get:

$\small{\int{\left[\sqrt{(\sqrt{5})^2 - (x+2)^2} \right]dx}}$

$\small{~=~~\frac{x+2}{2}\sqrt{(\sqrt{5})^2 - (x+2)^2} ~+~\frac{(\sqrt{5})^2}{2} \sin^{-1}\frac{x+2}{\sqrt{5}}~+~\rm{C}}$

$\small{\Rightarrow \int{\left[\sqrt{1-4x-x^2} \right]dx}}$

$\small{~=~\frac{x+2}{2}\sqrt{1-4x-x^2}~+~\frac{5}{2} \sin^{-1}\frac{x+2}{\sqrt{5}}~+~\rm{C}}$

---

The link below gives a few more solved examples:

Exercise 23.7

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We have completed a discussion on the three standard integrals. In the next section, we will see definite integrals.

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23.19 - Derivation of Some Standard Integrals

In the previous section, we completed a discussion on the method of integration by parts. In this section, we will see three standard integrals.

I. The integral $\bf{\int{\left[\sqrt{x^2 - a^2} \right]dx}}$

This can be calculated as follows:

1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\sqrt{x^2 - a^2}}$

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

2. Finding A:

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

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\sqrt{x^2 - a^2} \, \left(x\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{x}{\sqrt{x^2-a^2}}}$

(The reader must write all steps for this differentiation)

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{x}{\sqrt{x^2-a^2}}\,\left(x \right)  \big]dx}~=~\int{\big[\frac{x^2}{\sqrt{x^2-a^2}}  \big]dx}}$

• This can be rearranged as shown below:

$\small{\int{\big[\frac{x^2}{\sqrt{x^2-a^2}}  \big]dx}~=~\int{\big[\frac{x^2 - a^2 + a^2}{\sqrt{x^2-a^2}}  \big]dx}~=~\int{\big[\frac{x^2 - a^2}{\sqrt{x^2-a^2}}~+~\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx}}$

$\small{~=~\int{\big[\sqrt{x^2-a^2}~+~\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx}}$

$\small{~=~\int{\big[\sqrt{x^2-a^2} \big]dx}~+~\int{\big[\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx}}$

• This is the second term

6. So we get:

$\small{\int{\left[\sqrt{x^2 - a^2} \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[x \sqrt{x^2 - a^2} \big]~-~\bigg[\int{\big[\sqrt{x^2-a^2} \big]dx}~+~\int{\big[\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx} \bigg]}$

$\small{~=~x \sqrt{x^2 - a^2} ~-~\int{\big[\sqrt{x^2-a^2} \big]dx}~-~\int{\big[\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx} }$

$\small{\Rightarrow 2\int{\big[\sqrt{x^2-a^2} \big]dx}~=~x \sqrt{x^2 - a^2} ~-~\int{\big[\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx} }$

$\small{\Rightarrow \int{\big[\sqrt{x^2-a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 - a^2} ~-~\frac{a^2}{2} \int{\big[\frac{1}{\sqrt{x^2-a^2}}  \big]dx} }$

• The last term in the R.H.S can be calculated using formula IV that we saw in section 23.8. Thus we get:

$\small{\int{\big[\sqrt{x^2-a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 - a^2} ~-~\frac{a^2}{2} \log \left|x+\sqrt{x^2 - a^2} \right|~+~\rm{C}}$

II. The integral $\bf{\int{\left[\sqrt{x^2 + a^2} \right]dx}}$

This can be calculated as follows:

1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\sqrt{x^2 + a^2}}$

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

2. Finding A:

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

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\sqrt{x^2 + a^2} \, \left(x\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{x}{\sqrt{x^2+a^2}}}$

(The reader must write all steps for this differentiation)

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{x}{\sqrt{x^2+a^2}}\,\left(x \right)  \big]dx}~=~\int{\big[\frac{x^2}{\sqrt{x^2+a^2}}  \big]dx}}$

• This can be rearranged as shown below:

$\small{\int{\big[\frac{x^2}{\sqrt{x^2+a^2}}  \big]dx}~=~\int{\big[\frac{x^2 + a^2 - a^2}{\sqrt{x^2+a^2}}  \big]dx}~=~\int{\big[\frac{x^2 + a^2}{\sqrt{x^2-a^2}}~-~\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx}}$

$\small{~=~\int{\big[\sqrt{x^2+a^2}~-~\frac{ a^2}{\sqrt{x^2+a^2}}  \big]dx}}$

$\small{~=~\int{\big[\sqrt{x^2+a^2} \big]dx}~-~\int{\big[\frac{ a^2}{\sqrt{x^2+a^2}}  \big]dx}}$

• This is the second term

6. So we get:

$\small{\int{\left[\sqrt{x^2 + a^2} \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[x \sqrt{x^2 + a^2} \big]~-~\bigg[\int{\big[\sqrt{x^2+a^2} \big]dx}~-~\int{\big[\frac{ a^2}{\sqrt{x^2+a^2}}  \big]dx} \bigg]}$

$\small{~=~x \sqrt{x^2 + a^2} ~-~\int{\big[\sqrt{x^2+a^2} \big]dx}~+~\int{\big[\frac{ a^2}{\sqrt{x^2+a^2}}  \big]dx} }$

$\small{\Rightarrow 2\int{\big[\sqrt{x^2+a^2} \big]dx}~=~x \sqrt{x^2 + a^2} ~+~\int{\big[\frac{ a^2}{\sqrt{x^2+a^2}}  \big]dx} }$

$\small{\Rightarrow \int{\big[\sqrt{x^2+a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 + a^2} ~+~\frac{a^2}{2} \int{\big[\frac{1}{\sqrt{x^2+a^2}}  \big]dx} }$

• The last term in the R.H.S can be calculated using formula VI that we saw in section 23.8. Thus we get:

$\small{ \int{\big[\sqrt{x^2+a^2} \big]dx}~=~\frac{x}{2}\sqrt{x^2 + a^2} ~+~\frac{a^2}{2} \log \left|x+\sqrt{x^2 + a^2} \right|~+~\rm{C}}$

III. The integral $\bf{\int{\left[\sqrt{a^2 - x^2} \right]dx}}$

This can be calculated as follows:

1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{\sqrt{a^2 - x^2}}$

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

2. Finding A:

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

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\sqrt{a^2 - x^2} \, \left(x\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{-x}{\sqrt{a^2-x^2}}}$

(The reader must write all steps for this differentiation)

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{-x}{\sqrt{a^2-x^2}}\,\left(x \right)  \big]dx}~=~\int{\big[\frac{-x^2}{\sqrt{a^2-x^2}}  \big]dx}}$

• This can be rearranged as shown below:

$\small{\int{\big[\frac{-x^2}{\sqrt{a^2-x^2}}  \big]dx}~=~\int{\big[\frac{a^2 - x^2 - a^2}{\sqrt{a^2-x^2}}  \big]dx}~=~\int{\big[\frac{a^2 - x^2}{\sqrt{a^2-x^2}}~-~\frac{ a^2}{\sqrt{a^2-x^2}}  \big]dx}}$

$\small{~=~\int{\big[\sqrt{a^2-x^2}~-~\frac{ a^2}{\sqrt{a^2-x^2}}  \big]dx}}$

$\small{~=~\int{\big[\sqrt{a^2-x^2} \big]dx}~-~\int{\big[\frac{ a^2}{\sqrt{a^2-x^2}}  \big]dx}}$

• This is the second term

6. So we get:

$\small{\int{\left[\sqrt{a^2 - x^2} \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[x\sqrt{a^2 - x^2} \big]~-~\bigg[\int{\big[\sqrt{a^2-x^2} \big]dx}~-~\int{\big[\frac{ a^2}{\sqrt{a^2-x^2}}  \big]dx} \bigg]}$

$\small{~=~x\sqrt{a^2 - x^2} ~-~\int{\big[\sqrt{a^2-x^2} \big]dx}~+~\int{\big[\frac{ a^2}{\sqrt{a^2-x^2}}  \big]dx} }$

$\small{\Rightarrow 2\int{\big[\sqrt{a^2-x^2} \big]dx}~=~x\sqrt{a^2 - x^2} ~+~\int{\big[\frac{ a^2}{\sqrt{x^2-a^2}}  \big]dx} }$

$\small{\Rightarrow \int{\big[\sqrt{a^2-x^2} \big]dx}~=~\frac{x}{2}\sqrt{a^2 - x^2} ~+~\frac{a^2}{2} \int{\big[\frac{1}{\sqrt{a^2-x^2}}  \big]dx} }$

• The last term in the R.H.S can be calculated using formula V that we saw in section 23.8. Thus we get:

$\small{ \int{\big[\sqrt{a^2-x^2} \big]dx}~=~\frac{x}{2}\sqrt{a^2 - x^2} ~+~\frac{a^2}{2} \sin^{-1}\frac{x}{a}~+~\rm{C}}$

---

Now we will see some solved examples.

Solved example 23.61
Find $\small{\int{\left[\sqrt{x^2 +2x+ 5} \right]dx}}$
Solution:
1. The given integral can be rearranged as shown below:

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

2. Put u = x+1 Then du/dx = 1 ⇒ du = dx

So the given integral becomes: $\small{\int{\left[\sqrt{u^2+ 4} \right]du}}$

3. Now we can use the standard integral:

$\bf{\int{\left[\sqrt{x^2 + a^2} \right]dx}~=~\frac{x}{2}\sqrt{x^2 + a^2} ~+~\frac{a^2}{2} \log \left|x+\sqrt{x^2 + a^2} \right|~+~\rm{C}}$

Here x = u and a = 2

• So we get:

$\small{\int{\left[\sqrt{u^2 + 2^2} \right]du}~=~\frac{u}{2}\sqrt{u^2 + 2^2} ~+~\frac{2^2}{2} \log \left|u+\sqrt{u^2 + 2^2} \right|~+~\rm{C}}$

4. Substituting for u, we get:

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

$\small{\Rightarrow \int{\left[\sqrt{x^2 + 2x + 5} \right]dx}~=~\frac{x+1}{2}\sqrt{x^2 + 2x + 5} ~+~2 \log \left|(x+1)+\sqrt{x^2 + 2x + 5} \right|~+~\rm{C}}$

Solved Example 23.62
Find $\small{\int{\left[\sqrt{x^2 +4x+ 6} \right]dx}}$
Solution:
1. The given integral can be rearranged as shown below:

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

2. Put u = x+2 Then du/dx = 1 ⇒ du = dx

So the given integral becomes: $\small{\int{\left[\sqrt{u^2+ (\sqrt{2})^2} \right]du}}$

3. Now we can use the standard integral:

$\bf{\int{\left[\sqrt{x^2 + a^2} \right]dx}~=~\frac{x}{2}\sqrt{x^2 + a^2} ~+~\frac{a^2}{2} \log \left|x+\sqrt{x^2 + a^2} \right|~+~\rm{C}}$

Here x = u and a = √2

• So we get:

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

4. Substituting for u, we get:

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

$\small{\Rightarrow \int{\left[\sqrt{x^2 + 4x + 6} \right]dx}~=~\frac{x+2}{2}\sqrt{x^2 + 4x + 6} ~+~ \log \left|x+2+\sqrt{x^2 + 4x + 6} \right|~+~\rm{C}}$

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We have seen the three standard integrals. In the next section, we will see a few more solved examples.

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23.18 - Integration by Parts - Special case

In the previous section, we saw the method of Integration by parts. We saw some solved examples also. In this section, we will see a special case of this method.

Integral of the type $\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$

This can be calculated in 3 steps:

1. We have: $\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$

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

• There are two terms. We will denote the first one as I₁.

2. Let us calculate I₁:

(i) Assigning first and second functions:

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

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

(ii) Finding A:

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

(iii) $\small{\big[f(x) \left(A \right) \big]~=~\big[f(x) \, \left(e^x \right) \big]}$

• This is the first term.

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

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

• This is second term.

(vi) So we get:

I₁ = $\small{\text{First term - Second term}}$

$\small{~=~\big[f(x) \, \left(e^x \right) \big]~-~\bigg[\int{\big[f'(x)\,\left(e^x \right)  \big]dx} \bigg]}$

3. So from (1), we get:

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

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

= I₁ + $\small{\int{\left[e^x\,f'(x)\right]dx}}$

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

= $\small{f(x) \, \left(e^x \right)~+~\rm{C}}$

---

Let us see some solved examples:

Solved Example 23.57
Find $\small{\int{\big[e^x \left[\sin x + \cos x \right]\big]dx}}$
Solution:
1. Let f(x) = $\small{\sin x}$

• Then $\small{f'(x)~=~\cos x}$

2. So the given integral is of the form:
$\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$

3. Then the result is:

$\small{f(x) \, \left(e^x \right)~+~\rm{C}}$

$\small{~=~e^x \, \sin x~+~\rm{C}}$

Solved Example 23.58
Find $\small{\int{\left[e^x\,\sec x (1 + \tan x) \right]dx}}$
Solution:
1. The given integral can be written as:

$\small{\int{\big[e^x \left[\sec x + \sec x \tan x \right]\big]dx}}$

Let f(x) = $\small{\sec x}$

• Then $\small{f'(x)~=~\sec x \tan x}$

2. So the given integral is of the form:
$\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$

3. Then the result is:

$\small{f(x) \, \left(e^x \right)~+~\rm{C}}$

$\small{~=~e^x \, \sec x~+~\rm{C}}$

Solved Example 23.59
Find $\small{\int{\left[\frac{x e^x}{(1+x)^2} \right]dx}}$
Solution:
1. The $\small{\left[\frac{x}{(1+x)^2} \right]}$ portion can be rearranged as shown below:

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

2. Now, $\small{\frac{(-1)}{(1+x)^2}}$ is the derivative of $\small{\frac{1}{1+x}}$.

(The reader may verify this by doing the differentiation)

3. So the given integral is of the form:

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

Where $\small{f(x) = \frac{1}{1+x}}$

3. Then the result is:

$\small{f(x) \, \left(e^x \right)~+~\rm{C}}$

$\small{~=~e^x \, \left[\frac{1}{1+x} \right]~+~\rm{C}}$

Solved example 23.60
Find $\small{(i)~\int{\bigg[e^x \left[\tan^{-1}x \,+\,\frac{1}{1+x^2} \right]\bigg]dx}~~~~~~(ii)~\int{\left[\frac{(x^2 + 1)e^x}{(x+1)^2} \right]dx}}$
Solution:
Part (i):
1. Let f(x) = $\small{\tan^{-1} x}$
• Then $\small{f'(x)~=~\frac{1}{1 + x^2}}$

2. So the given integral is of the form:
$\small{\int{\big[e^x \left[f(x)\,+\,f'(x)\right]\big]dx}}$

3. Then the result is:

$\small{f(x) \, \left(e^x \right)~+~\rm{C}}$

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

Part (ii):

1. The $\small{\left[\frac{(x^2 + 1)}{(x+1)^2} \right]}$ portion can be rearranged as shown below:

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

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

2. Now, $\small{\frac{2}{(x+1)^2}}$ is the derivative of $\small{\frac{x-1}{x+1}}$.

(The reader may verify this by doing the differentiation)

3. So the given integral is of the form:

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

Where $\small{f(x) = \frac{x-1}{x+1}}$

3. Then the result is:

$\small{f(x) \, \left(e^x \right)~+~\rm{C}}$

$\small{~=~e^x \, \left[\frac{x-1}{x+1} \right]~+~\rm{C}}$

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The link below gives a few more solved examples:

Exercise 23.6

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We have completed a discussion on integration by parts. In the next section, we will see some standard integrals.

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23.17 - Solved Examples on Integration by Parts

In the previous section, we saw the basic details about Integration by parts. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved Example 23.45
Find $\small{\int{\left[x \, \log 2x \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = log 2x

   ♦ Let second function be: g(x) = x

2. Finding A:

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

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\log 2x \, \left(\frac{x^2}{2} \right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{1}{x}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{1}{x}\,\left(\frac{x^2}{2} \right)  \big]dx}~=~\frac{x^2}{4}}$

• This is the second term.

6. So we get:

$\small{\int{\left[x \, \log 2x \right]dx}~=~\text{First term - Second term}~=~\log 2x \, \left(\frac{x^2}{2} \right)~-~\frac{x^2}{4}~+~\rm{C}}$

Solved Example 23.46
Find $\small{\int{\left[x^2 \, \log x \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = log x

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

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[x^2\right]dx}~=~\frac{x^3}{3}}$

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\log x \, \left(\frac{x^3}{3} \right) \big]}$

• This is the first term.

4. $\small{~f'(x)~=~\frac{1}{x}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{1}{x}\,\left(\frac{x^3}{3} \right)  \big]dx}~=~\frac{x^3}{9}}$

• This is the second term.

6. So we get:

$\small{\int{\left[x^2 \, \log x \right]dx}~=~\text{First term - Second term}~=~\log x \, \left(\frac{x^3}{3} \right)~-~\frac{x^3}{9}~+~\rm{C}}$

Solved Example 23.47
Find $\small{\int{\left[x \, \sin^{-1} x \right]dx}}$
Solution:
1. Assigning first and second functions:

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

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

2. Finding A:

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

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

• This is the first term.

4. $\small{f'(x)~=~\frac{1}{\sqrt{1-x^2}}}$

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

• This is the second term. But is also a product. We will do it as a sub group:

*** Beginning of sub group ***

We want $\small{\frac{1}{2}\int{\big[\frac{x^2}{\sqrt{1-x^2}}  \big]dx}}$

(i) Let x = sin u

$\small{\Rightarrow \frac{dx}{du}\,=\,\cos u \Rightarrow\frac{dx}{\cos u}\,=\,du}$

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

(ii) So we want:

$\small{\frac{1}{2}\int{\big[\frac{\sin^2 u}{\cos u}  \big]dx}\,=\,\frac{1}{2}\int{\big[\sin^2 u  \big]du}}$

(iii) This can be calculated as:

$\small{\frac{1}{2}\int{\big[\sin^2 u  \big]du}\,=\,\frac{1}{2}\int{\big[\frac{1 - \cos 2u}{2}  \big]du}\,=\,\frac{1}{4}\int{\big[1 - \cos 2u  \big]du}}$

$\small{~=~\frac{u}{4}\,-\,\frac{\sin 2u}{8}~=~\frac{u}{4}\,-\,\frac{2 \sin u \cos u}{8}}$

$\small{~=~\frac{u}{4}\,-\,\frac{\sin u \cos u}{4}~=~\frac{\sin^{-1}x}{4}\,-\,\frac{x \sqrt{1 - x^2}}{4}}$

*** End of sub group ***

6. So we get:

$\small{\int{\left[x \, \sin^{-1} x \right]dx}~=~\text{First term - Second term}}$

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

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

Solved Example 23.48
Find $\small{\int{\left[x \, \tan^{-1} x \right]dx}}$
Solution:
1. Assigning first and second functions:

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

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

2. Finding A:

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

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

• This is the first term.

4. $\small{f'(x)~=~\frac{1}{1+x^2}}$

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

• This is the second term. But is also a product. We will do it as a sub group:

*** Beginning of sub group ***

We want $\small{\frac{1}{2}\int{\big[\frac{x^2}{1+x^2}  \big]dx}}$

(i) Let x = tan u

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

Also, $\small{~1 + x^2\,=\,\sec^2 u}$

(ii) So we want:

$\small{\frac{1}{2}\int{\big[\frac{\tan^2 u}{\sec^2 u}  \big]dx}\,=\,\frac{1}{2}\int{\big[\tan^2 u  \big]du}}$

(iii) This can be calculated as:

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

$\small{\,=\,\frac{x}{2}\,-\,\frac{\tan^{-1}x}{2}}$

*** End of sub group ***

6. So we get:

$\small{\int{\left[x \, \tan^{-1} x \right]dx}~=~\text{First term - Second term}}$

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

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

Solved Example 23.49
Find $\small{\int{\left[x \, \cos^{-1} x \right]dx}}$
Solution:
1. Assigning first and second functions:

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

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

2. Finding A:

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

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\cos^{-1}x \, \left(\frac{x^2}{2}\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{-1}{\sqrt{1-x^2}}}$

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

• This is the second term. But is also a product. We will do it as a sub group:

*** Beginning of sub group ***

We want $\small{\frac{-1}{2}\int{\big[\frac{x^2}{\sqrt{1-x^2}}  \big]dx}}$

(i) Let x = cos u

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

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

(ii) So we want:

$\small{\frac{-1}{2}\int{\big[\frac{\cos^2 u}{\sin u}  \big]dx}\,=\,\frac{1}{2}\int{\big[\cos^2 u  \big]du}}$

(iii) This can be calculated as:

$\small{\frac{1}{2}\int{\big[\cos^2 u  \big]du}\,=\,\frac{1}{2}\int{\big[\frac{1 + \cos 2u}{2}  \big]du}\,=\,\frac{1}{4}\int{\big[1 + \cos 2u  \big]du}}$

$\small{~=~\frac{u}{4}\,+\,\frac{\sin 2u}{8}~=~\frac{u}{4}\,+\,\frac{2 \sin u \cos u}{8}}$

$\small{~=~\frac{u}{4}\,+\,\frac{\sin u \cos u}{4}~=~\frac{\cos^{-1}x}{4}\,+\,\frac{x \sqrt{1 - x^2}}{4}}$

*** End of sub group ***

6. So we get:

$\small{\int{\left[x \, \cos^{-1} x \right]dx}~=~\text{First term - Second term}}$

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

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

Solved Example 23.50
Find $\small{\int{\left[\sin^{-1}x \right]dx}}$
Solution:
1. Assigning first and second functions:

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

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

2. Finding A:

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

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\sin^{-1}x \, \left(x\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{1}{\sqrt{1-x^2}}}$

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

• This is second term.

(The reader may write all steps involved in this integration process)

6. So we get:

$\small{\int{\left[\sin^{-1} x \right]dx}~=~\text{First term - Second term}}$

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

$\small{~=~x \sin^{-1}x~+~\sqrt{1-x^2}\,+\,\rm{C}}$

Solved Example 23.51
Find $\small{\int{\left[(\sin^{-1}x)^2 \right]dx}}$
Solution:
1. Assigning first and second functions:

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

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

2. Finding A:

$\small{A~=~\int{\left[g(x) \right]dx}~=~\int{\left[\sin^{-1}x \right]dx}~=~x \sin^{-1}x~+~\sqrt{1-x^2}}$

(See solved example 23.50 above)

3. $\small{\big[f(x) \left(A \right) \big]~=~\sin^{-1}x \big[x \sin^{-1}x~+~\sqrt{1-x^2} \big]}$

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

• This is the first term.

4. $\small{f'(x)~=~\frac{1}{\sqrt{1-x^2}}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{1}{\sqrt{1-x^2}}\,\left(x \sin^{-1}x~+~\sqrt{1-x^2} \right)  \big]dx}}$

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

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

(See solved example 23.38 of the previous section)

• This is second term.

6. So we get:

$\small{\int{\left[(\sin^{-1} x)^2 \right]dx}~=~\text{First term - Second term}}$

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

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

Solved Example 23.52
Find $\small{\int{\left[\frac{x \cos^{-1} x}{\sqrt{1 - x^2}} \right]dx}}$
Solution:
1. Assigning first and second functions:

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

   ♦ Let second function be: g(x) = $\small{\frac{x }{\sqrt{1 - x^2}} }$

2. Finding A:

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

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[(-1)(\cos^{-1}x) \,\sqrt{1-x^2} \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{-1}{\sqrt{1 - x^2}}}$

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

• This is the second term.

6. So we get:

$\small{\int{\left[\frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \right]dx}~=~\text{First term - Second term}}$

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

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

Solved Example 23.53
Find $\small{\int{\left[x \sec^2 x \right]dx}}$
Solution:
$\small{\int{\left[x \sec^2 x \right]dx}}$

1. Assigning first and second functions:

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

   ♦ Let second function be: g(x) = $\small{\sec^2 x }$

2. Finding A:

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

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[x \tan x \big]}$

• This is the first term.

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

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[1\,\left(\tan x \right)  \big]dx}~=~\log \left|\sec x \right| }$

• This is the second term.

6. So we get:

$\small{\int{\left[\frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[x \tan x \big]~-~\big[\log \left|\sec x \right| \big]~+~\rm{C}}$

$\small{~=~x \tan x ~+~\log \left|\frac{1}{\sec x} \right| ~+~\rm{C}}$

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

Solved Example 23.54
Find $\small{\int{\left[\tan^{-1}x \right]dx}}$
Solution:
1. Assigning first and second functions:

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

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

2. Finding A:

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

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\tan^{-1}x \, \left(x\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{1}{1+x^2}}$

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

• This is second term.

(The reader may write all steps involved in this integration process)

6. So we get:

$\small{\int{\left[\tan^{-1} x \right]dx}~=~\text{First term - Second term}}$

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

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

Solved Example 23.55
Find $\small{\int{\left[x (\log x)^2 \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = $\small{(\log x)^2}$

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

2. Finding A:

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

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[(\log x)^2 \, \left(\frac{x^2}{2}\right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{2 \log \left|x \right|}{x}}$

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

$\small{~=~\int{\big[x\log \left|x \right|  \big]dx}~=~\log x \, \left(\frac{x^2}{2} \right)~-~\frac{x^2}{4}}$

(See solved example 23.44 of the previous section)

• This is second term.

6. So we get:

$\small{\int{\left[x (\log x)^2 \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\big[(\log x)^2 \, \left(\frac{x^2}{2}\right) \big]~-~\big[\log x \, \left(\frac{x^2}{2} \right)~-~\frac{x^2}{4} \big]}$

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

Solved Example 23.56
Find $\small{\int{\left[(x^2 + 1) \log x \right]dx}}$
Solution:
1. Assigning first and second functions:

   ♦ Let first function be: f(x) = log x

   ♦ Let second function be: g(x) = $\small{x^2 + 1}$

2. Finding A:

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

3. $\small{\big[f(x) \left(A \right) \big]~=~\big[\log x \, \left(\frac{x^3}{3}~+~x \right) \big]}$

• This is the first term.

4. $\small{f'(x)~=~\frac{1}{x}}$

5. $\small{\int{\big[f'(x)\,\left(A \right)  \big]dx}~=~\int{\big[\frac{1}{x}\,\left(\frac{x^3}{3}~+~x \right)  \big]dx}~=~\frac{x^3}{9}~+~x}$

• This is the second term.

6. So we get:

$\small{\int{\left[(x^2 + 1) \log x \right]dx}~=~\text{First term - Second term}}$

$\small{~=~\log x \, \left(\frac{x^3}{3}~+~x \right)~-~\frac{x^3}{9}~-~x~+~\rm{C}}$

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We have seen the method of integration by parts. In the next section, we will see a special case in this method.

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