A) \[\frac{\pi }{2}\]
B) \[\frac{\pi }{6}\]
C) \[\frac{\pi }{3}\]
D) \[\frac{\pi }{5}\]
Correct Answer: B
Solution :
Let \[I=\int_{0}^{\pi /2}{\frac{\cos \theta }{\sqrt{4-{{\sin }^{2}}\theta }}d\theta }\] Put \[\sin \theta =t\]\[\Rightarrow \]\[\cos \theta d\theta =dt\] \[\therefore \] \[I=\int_{0}^{1}{\frac{dt}{\sqrt{4-{{t}^{2}}}}}\] \[=\left[ {{\sin }^{-1}}\frac{t}{2} \right]_{0}^{1}={{\sin }^{-1}}\frac{1}{2}\] \[=\frac{\pi }{6}\]
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