Question

Value of $\int_{\pi / 6}^{\pi / 3} \frac{1}{1+\sqrt{\cot x}} d x$ is

• A. $\frac{\pi}{6}$
• B. $\frac{\pi}{12}$
• C. $\frac{12}{\pi}$
• D. None of these

Answer 1

Answer: (B)

$
I=\int_{\pi / 6}^{\pi / 3} \frac{1}{1+\sqrt{\cot x}} d x=\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \ldots \text { (i) }
$
Then, $I=\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}+\sqrt{\cos \left(\frac{\pi}{2}-x\right)}} d x$
$
\Rightarrow I=\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x \ldots \text { (ii) }
$
Adding (i) and (ii), we get
$
2 I=\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x
$
$
\Rightarrow 2 I=\int_{\pi / 6}^{\pi / 3} 1 . d x=[x]_{\pi / 6}^{\pi / 3}
$
$
=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6} \Rightarrow I=\pi / 12
$