Question

${\int }_0^a\sqrt{\frac{a-x}{x}}dx$ =

• A. $\frac{\pi a}{2}$
• B. $\frac{\pi a}{4}$
• C. $\frac{\pi a^2}{4}$
• D. $\frac{a}{2}$

Answer 1

Answer: (A)

Let $I={\int }_0^a\sqrt{\frac{a-x}{x}}dx$
Put $x=a{\sin }^2\theta$
$\Rightarrow$ $dx=2a\sin \theta \cos \theta d\theta =a\sin 2\theta d\theta$
$\therefore$ $I={\int }_0^{\pi /2}\sqrt{\frac{a-a{\sin }^2\theta }{a{\sin }^2\theta }}.a\sin 2\theta d\theta$
$=a{\int }_0^{\pi /2}\frac{\cos \theta }{\sin \theta }.2\sin \theta \cos \theta d\theta$
$=2a{\int }_0^{\pi /2}{\cos }^2\theta d\theta$
$=2a\left[\frac{2-1}{2}.\frac{\pi }{2}\right]$
[using walli's formula] $=2a\times \frac{\pi }{4}$
$=\frac{\pi a}{2}$