Question

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

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

Answer 1

Answer: (B)

Let $I=\underset{0}{\overset{\pi /2}{\int }}\frac{\cos \theta }{\sqrt{4-{\sin }^2\theta }}d\theta$
Put $\sin \theta =t$
$\Rightarrow \cos \theta d\theta =dt$
$\therefore I=\underset{0}{\overset{1}{\int }}\frac{dt}{\sqrt{4-t^2}}$
$={\left[{\sin }^{-1}\frac{t}{2}\right]}_0^1={\sin }^{-1}\frac{1}{2}$
$=\frac{\pi }{6}$