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} $