A) \[\frac{\pi a}{2}\]
B) \[\frac{\pi a}{4}\]
C) \[\frac{\pi {{a}^{2}}}{4}\]
D) \[\frac{a}{2}\]
Correct Answer: A
Solution :
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] \[=2a\times \frac{\pi }{4}\] \[=\frac{\pi a}{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
