A) \[\frac{a}{2}\]
B) \[\frac{a}{4}\]
C) \[\frac{\pi a}{2}\]
D) \[\frac{\pi a}{4}\]
Correct Answer: C
Solution :
Let \[I=\int_{0}^{a}{\sqrt{\frac{a-x}{x}}}\,dx\] Put \[x=a{{\sin }^{2}}\theta \] and \[dx=2a\sin \theta \cos \theta d\theta \] \[\therefore \]\[I=\int_{0}^{\pi /2}{\sqrt{\frac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }}}\cdot 2a\,\sin \theta \,\cos \theta d\theta \] \[=2a\int_{0}^{\pi /2}{{{\cos }^{2}}\theta \,d\theta }\] \[=2a\times \frac{1}{2}\times \frac{\pi }{2}=\frac{\pi a}{2}\]
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