Q. If $b > a$, then $\int_{a}^{b}\frac{dx}{\sqrt{(x - a)(b - x)}}$ is equal to

A) $\frac{\pi}{2}$
B) $\pi$
C) $\frac{\pi}{2}(b - a)$
D) $\frac{\pi}{4}(b - a)$
Explanation

(b)
$\int_{a}^{b}\frac{dx}{\sqrt{(x - a)(b - x)}} = \int_{a}^{b}\frac{1}{\sqrt{- x^{2} + (a + b)x - ab}}\ dx$
$= \int_{a}^{b}\frac{1}{\sqrt{\left( \frac{b - a}{2} \right)^{2} - \left( x - \frac{a + b}{2} \right)^{2}}}dx$
$= \left\lbrack \sin^{- 1}\left( \frac{x - \frac{a + b}{2}}{\frac{b - a}{2}} \right) \right\rbrack_{a}^{b}$
$= \sin^{- 1}1 - \sin^{- 1}{( - 1)}$
$= \frac{\pi}{2} + \frac{\pi}{2} = \pi$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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