Question

The value of $\int^{^a}_{0}\sqrt{\frac{a-x}{x}}dx$ is

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

Answer 1

The Correct Option is C

Let x = cos$^2$, $\theta$so that dx = - 2a sin $\theta$ cos $\theta$ d$\theta$
Now $\int^{^a}_{0}\sqrt{\frac{a-x}{x}}dx=\int^{0}_{\frac{\pi}{2}}\sqrt{\frac{a-a cos^{2} \theta}{a cos^{2} \theta}}$
$\quad\quad\quad\quad\quad\left(-a sin \theta cos \theta \right)d \theta $
$\quad\quad \quad \quad \quad \quad \quad \quad \left[\because \theta=\frac{\pi}{2} at x=0;\theta=0 at x =a \right]$
$=a\int^{^{\pi/2}}_{_0}\sqrt{\frac{1-cos^{2} \theta}{cos^{2} \theta }}$ 2 sin $\theta$ cos $\theta$ d $\theta $
$\quad$$\quad$$\quad$$\quad$$\left[\because \int^{0}_{t} f \left(x\right)dx=-\int^{0}_{t} f \left(x\right)dx\right]$
$=a\int^{\pi/2}_{0}2\frac{sin \theta }{cos \theta }.sin \theta cos \theta d\theta $
$=a\int^{\pi/2}_{0} 2sin^{2} \theta d\theta =a\int ^{\pi/ 2}_{0} \left(1-cos 2 \theta \right)d \theta $
$\quad\quad\quad\quad\left[\because cos 2\theta =1-2sin^{2} \theta \right]$
$=a\left[\theta -\frac{sin 2\theta }{2}\right]^{^{\pi/2}}_{_{_0}}$
$=a\left[\left(\frac{\pi}{2}-\frac{sin \pi}{2}\right)-\left(0-0\right)\right]$
$=a\left(\frac{\pi}{2}-0\right)=\frac{a\pi}{2}$