A) \[4a\]
B) \[-4a\]
C) \[2a\]
D) \[a\]
Correct Answer: C
Solution :
Let\[I=\int_{0}^{a}{\sqrt{\frac{a}{a-x}}dx}\] Put \[x=a{{\cos }^{2}}\theta \] \[\Rightarrow \] \[dx=-2a\cos \theta \sin \theta d\theta \] \[\therefore \] \[I=\int_{\pi /2}^{0}{\sqrt{\frac{a}{a-a{{\cos }^{2}}\theta }}.(-2a\cos \theta \sin \theta )d\theta }\] \[=-\int_{\pi /2}^{0}{\frac{1}{\sqrt{{{\sin }^{2}}\theta }}}.2a\cos \theta .\sin \theta d\theta \] \[=-\int_{\pi /2}^{0}{2a\cos \theta }d\theta \] \[=-2a[\sin \theta ]_{\pi /2}^{0}\] \[=-2a\left( \sin 0-\sin \frac{\pi }{2} \right)\]
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