Chapter :- method of substitution

1.	Chapter :- method of substitution
Answer» Given to evaluate, $\displaystyle\longrightarrow I=\int\dfrac{dx}{(x-b)^<klux>3</klux>(x-a)^<klux>2</klux>}\quad\quad\dots(<klux>1</klux>)$ 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), $\displaystyle\longrightarrow\dfrac{x-a}{x-b}=\dfrac{(b-a)\sin^2\theta}{(a-b)\cos^2\theta}$ $\displaystyle\longrightarrow\tan^2\theta=\dfrac{x-a}{b-x}\quad\quad\dots(<klux>5</klux>)$ 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»

Given to evaluate,

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

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),

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

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

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,

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

Chapter :- method of substitution

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Chapter :- method of substitution