Question

chapter :- method of substitution​

Answer 1

Given to evaluate,

$\displaystyle\longrightarrow I=\int\dfrac{dx}{(x-b)^3(x-a)^2}\quad\quad\dots(1)$

Substitute,

$\displaystyle\longrightarrow x=a\cos^2\theta+b\sin^2\theta\quad\quad\dots(2)$

$\displaystyle\longrightarrow dx=\left(-2a\sin\theta\cos\theta+2b\sin\theta\cos\theta\right)\ d\theta$

$\displaystyle\longrightarrow dx=2(b-a)\sin\theta\cos\theta\ d\theta$

Now, adding $a\sin^2\theta$ to both sides of (2),

$\displaystyle\longrightarrow x+a\sin^2\theta=a\cos^2\theta+b\sin^2\theta+a\sin^2\theta$

Since $\sin^2\theta+\cos^2\theta=1,$

$\displaystyle\longrightarrow x+a\sin^2\theta=a+b\sin^2\theta$

$\displaystyle\longrightarrow x-a=(b-a)\sin^2\theta\quad\quad\dots(3)$

Similarly, adding $b\cos^2\theta$ to both sides of (2),

$\displaystyle\longrightarrow x+b\cos^2\theta=a\cos^2\theta+b\sin^2\theta+b\cos^2\theta$

$\displaystyle\longrightarrow x+b\cos^2\theta=a\cos^2\theta+b$

$\displaystyle\longrightarrow x-b=(a-b)\cos^2\theta\quad\quad\dots(4)$

Dividing (3) by (4),

$\displaystyle\longrightarrow\dfrac{x-a}{x-b}=\dfrac{(b-a)\sin^2\theta}{(a-b)\cos^2\theta}$

$\displaystyle\longrightarrow\tan^2\theta=\dfrac{x-a}{b-x}\quad\quad\dots(5)$

Thus (1) becomes,

$\displaystyle\longrightarrow I=\int\dfrac{2(b-a)\sin\theta\cos\theta}{\left[(a-b)\cos^2\theta\right]^3\left[(b-a)\sin^2\theta\right]^2}\ d\theta$

$\displaystyle\longrightarrow I=\dfrac{2(b-a)}{(a-b)^5}\int\dfrac{\sin\theta\cos\theta}{\cos^6\theta\sin^4\theta}\ d\theta$

$\displaystyle\longrightarrow I=-\dfrac{2}{(a-b)^4}\int\dfrac{1}{\cos^5\theta\sin^3\theta}\ d\theta$

Dividing both numerator and denominator of the integrand by $\cos^3\theta,$

$\displaystyle\longrightarrow I=-\dfrac{2}{(a-b)^4}\int\dfrac{\left(\dfrac{1}{\cos^3\theta}\right)}{\left(\dfrac{\cos^5\theta\sin^3\theta}{\cos^3\theta}\right)}\ d\theta$

$\displaystyle\longrightarrow I=-\dfrac{2}{(a-b)^4}\int\dfrac{\left(\dfrac{1}{\cos^3\theta}\right)}{\tan^3\theta\cos^5\theta}\ d\theta$

$\displaystyle\longrightarrow I=-\dfrac{2}{(a-b)^4}\int\dfrac{\left(\dfrac{1}{\cos^8\theta}\right)}{\tan^3\theta}\ d\theta$

$\displaystyle\longrightarrow I=-\dfrac{2}{(a-b)^4}\int\dfrac{\sec^8\theta}{\tan^3\theta}\ d\theta$

$\displaystyle\longrightarrow I=-\dfrac{2}{(a-b)^4}\int\dfrac{\left(1+\tan^2\theta\right)^3\sec^2\theta\ d\theta}{\tan^3\theta}\quad\quad\dots(6)$

Substitute,

$\longrightarrow u=\tan\theta$

$\longrightarrow du=\sec^2\theta\ d\theta$

Then (6) becomes,

$\displaystyle\longrightarrow I=-\dfrac{2}{(a-b)^4}\int\dfrac{\left(1+u^2\right)^3}{u^3}\ du$

$\displaystyle\longrightarrow I=-\dfrac{2}{(a-b)^4}\int\dfrac{1+3u^2+3u^4+u^6}{u^3}\ du$

$\displaystyle\longrightarrow I=-\dfrac{2}{(a-b)^4}\int\left(\dfrac{1}{u^3}+\dfrac{3}{u}+3u+u^3\right)\ du$

$\displaystyle\longrightarrow I=-\dfrac{2}{(a-b)^4}\left[-\dfrac{1}{2u^2}+3\ln|u|+\dfrac{3u^2}{2}+\dfrac{u^4}{4}\right]+C$

$\displaystyle\longrightarrow I=\dfrac{1}{(a-b)^4}\left[\dfrac{1}{u^2}-3\ln\left(u^2\right)-3u^2-\dfrac{u^4}{2}\right]+C$

Undoing substitution $u=\tan\theta,$

$\displaystyle\longrightarrow I=\dfrac{1}{(a-b)^4}\left[\dfrac{1}{\tan^2\theta}-3\ln\left(\tan^2\theta\right)-3\tan^2\theta-\dfrac{\tan^4\theta}{2}\right]+C$

From (5), we get,

$\displaystyle\longrightarrow\underline{\underline{I=\dfrac{1}{(a-b)^4}\left[\dfrac{3(x-a)}{x-b}-\dfrac{x-b}{x-a}-\dfrac{1}{2}\left(\dfrac{x-a}{x-b}\right)^2-3\ln\left|\dfrac{x-a}{b-x}\right|\right]+C}}$

Answer 2

$\rm \: \longrightarrow I=\int\dfrac{dx}{(x-b)^3(x-a)^2} - - - (i)$

Now,

To solve this integral,

Put,

$\rm \: \implies\:x=a\cos^2\theta+b\sin^2\theta$

On differentiating both sides, we get

$\rm : \implies \:dx=\left(-2a\sin\theta\cos\theta+2b\sin\theta\cos\theta\right)\ d\theta$

$\rm : \implies \:dx=2(b-a)\sin\theta\cos\theta\ d\theta$

Now,

$\rm : \implies \:x - a=a\cos^2\theta+b\sin^2\theta \: - a$

$\rm : \implies \:x - a= - a(1 - \cos^2\theta)+b\sin^2\theta \:$

$\rm : \implies \:x - a= - a\sin^2\theta+b\sin^2\theta \:$

$\rm : \implies \:x - a=(a - b)\sin^2\theta \: - - - (ii)$

Now,

$\rm : \implies \:x - b=a\cos^2\theta+b\sin^2\theta \: - b$

$\rm : \implies \:x - b=a\cos^2\theta -b (1 - \sin^2\theta) \:$

$\rm : \implies \:x - b=a\cos^2\theta -b \cos^2\theta\:$

$\rm : \implies \:x - b=(a -b)\cos^2\theta\: - - - (iii)$

Now, Substitute all these values in equation (1), we get

$\rm : \implies \:I = \int \: \dfrac{2(b-a)\sin\theta\cos\theta \: d \theta}{ {(a - b)}^{3} {cos}^{6}\theta {(a - b)}^{2} {sin}^{4} \theta }$

$\rm : \implies \:I=\dfrac{ - 2}{(a-b)^4}\int\dfrac{1}{\cos^5\theta\sin^3\theta}\ d\theta$

$\sf \: Dividing \: both \: numerator \: and \: denominator \: by \:  \cos^8\theta$

$\rm : \implies \:I=-\dfrac{2}{(a-b)^4}\int\dfrac{ \sec^2\theta}{\tan^3\theta}\ d\theta$

Now,

Substitute,

$\rm : \implies \: \tan\theta \: = y$

On differentiating both sides, we get

$\rm : \implies \: \sec^2\theta \: d \theta \: = \: dy$

$\rm : \implies \:I=-\dfrac{2}{(a-b)^4}\int \: \dfrac{dy}{ {y}^{3} }$

$\rm : \implies \:I=-\dfrac{2}{(a-b)^4}\int \: {y}^{ - 3} dy$

$\rm : \implies \:I=-\dfrac{2}{(a-b)^4} \: \dfrac{ {y}^{ - 2} }{ (- 2)} + c$

$\rm : \implies \:I=\dfrac{1}{ {y}^{2} (a-b)^4} \: + \: c$

$\rm : \implies \:I=\dfrac{1}{ {tan}^{2} \theta \: (a-b)^4} \: + \: c - - (iv)$

$\boxed{ \pink{ \rm : \implies \:Now, \: to \: find \: tan \: \theta}}$

Divide equation (ii) by equation (iii), we get

$\rm : \implies \:\dfrac{x - a}{x - b} = {tan}^{2} \theta$

Substitute this value in equation (iv), we get

$\rm : \implies \:I=\dfrac{ {(x - b)}^{2} }{ {(x - a)}^{2} (a-b)^4} \: + \: c$