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:
$\small{2\theta~=~\cos^{-1}x ~~\text{and}~~ \sin(2\theta)~=~\sqrt{1 - x^2}}$
$\small{~=~ \int{\left[\theta\right]dx}~=~\int{\left[\theta\right]}(-2 \sin(2 \theta) d \theta)~=~-2\int{\left[\theta \sin(2 \theta)\right]}d \theta}$
$\small{\int{\left[\theta \sin(2 \theta)\right]}d \theta}$ 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.
$\small{\frac{1}{(x^2 + 1)(x^2 + 4)}~=~\frac{(A_1 x + B_1) (x^2 + 4)~+~(A_2 x + B_2) (x^2 + 1)}{(x+1)(x^2 + 4)}}$
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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