Question

The value of $\int\limits_{0}^{\pi / 2} \frac{\cos \theta}{\sqrt{4-\sin ^{2} \theta}} d \theta$ is :

• A. $ \frac{\pi }{2} $
• B. $ \frac{\pi }{6} $
• C. $ \frac{\pi }{3} $
• D. $ \frac{\pi }{5} $

Answer 1

Answer: (B)

Let $I =\int\limits_{0}^{\pi / 2} \frac{\cos \theta}{\sqrt{4-\sin ^{2} \theta}} d \theta$
Put $\sin \theta =t$
$\Rightarrow \cos \theta d \theta=d t$
$\therefore I =\int\limits_{0}^{1} \frac{d t}{\sqrt{4-t^{2}}}$
$=\left[\sin ^{-1} \frac{t}{2}\right]_{0}^{1}=\sin ^{-1} \frac{1}{2}$
$=\frac{\pi}{6}$