Question

Value of ${\int }_{\pi /6}^{\pi /3}\frac{1}{1+\sqrt{\cot x}}dx$ is

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

Answer 1

Answer: (B)

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