Question

Evaluate $\int \frac{dx}{\sqrt{(x-a)(b-x)}}$.

• A. $I=2{\sin }^{-1}\sqrt{\frac{x-a}{(b-a)}}+C$
• B. $I=2{\cos }^{-1}\sqrt{\frac{x-a}{(b-a)}}+C$
• C. $I={\sin }^{-1}\sqrt{\frac{x-a}{(b-a)}}+C$
• D. $I=2{\sin }^{-1}\sqrt{\frac{x-b}{(a-b)}}+C$

Answer 1

The correct option is A $I=2{\sin }^{-1}\sqrt{\frac{x-a}{(b-a)}}+C$

$I=\int \frac{dx}{\sqrt{(x-a)(b-x)}}$
Put $x=a{\cos }^2\theta +b{\sin }^2\theta \Rightarrow x=a(1-{\sin }^2\theta )+b{\sin }^2\theta$

$\Rightarrow x-a={\sin }^2\theta (b-a)\Rightarrow \sin \theta =\sqrt{\frac{x-a}{b-a}}\Rightarrow \theta ={\sin }^{-1}\sqrt{\frac{x-a}{b-a}}$

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

$\Rightarrow dx=2(b-a)\sin \theta \cos \theta d\theta$
$\therefore I=\int \frac{2(b-a)\sin \theta \cos \theta d\theta }{\sqrt{(a{\cos }^2\theta +b{\sin }^2\theta -a)(b-a{\cos }^2\theta -b{\sin }^2\theta )}}$
$=2(b-a)\int \frac{\sin \theta \cos \theta d\theta }{\sqrt{(a{\cos }^2\theta +b{\sin }^2\theta -a)(b-a{\cos }^2\theta -b{\sin }^2\theta )}}$
$=2(b-a)\int \frac{\sin \theta \cos \theta d\theta }{\sqrt{(b{\sin }^2\theta -a{\sin }^2\theta )(b{\cos }^2\theta -a{\cos }^2\theta )}}$
$=2(b-a)\int \frac{\sin \theta \cos \theta d\theta }{(b-a)\sin \theta \cos \theta }$
$=2\int 1d\theta$
$=2\theta +C=2{\sin }^{-1}\sqrt{\frac{x-a}{(b-a)}}+C$