Question

$I={\int }_{\pi /6}^{\pi /3}\frac{1}{1+\sqrt{\left(\cot x\right)}}dx$

• A. $-\frac{\pi }{6}.$
• B. $\frac{\pi }{6}.$
• C. $-\frac{\pi }{12}.$
• D. $\frac{\pi }{12}.$

Answer 1

The correct option is D $\frac{\pi }{12}.$

$I={\int }_{\pi /6}^{\pi /3}\frac{\sqrt{\sin x}}{\sqrt{\left(\sin x\right)}+\sqrt{\left(\cos x\right)}}dx.$ ...(1)
Substitute $x=\pi /2-t\because a+b=\pi /6+\pi /3=\pi /2$
$\therefore I={\int }_{\pi /3}^{\pi /6}\frac{\sqrt{\cos t}}{\sqrt{\left(\cos t\right)}+\sqrt{\left(\sin t\right)}}(-dt)$
$={\int }_{\pi /6}^{\pi /3}\frac{\sqrt{\cos t}}{\sqrt{\left(\cos t\right)}+\sqrt{\left(\sin t\right)}}dt$
or
$I={\int }_{\pi /3}^{\pi /6}\frac{\sqrt{\cos x}dx}{\sqrt{\left(\cos x\right)}+\sqrt{\left(\sin x\right)}},$ ...(2)
Adding (1) and (2), we get
$2I={\int }_{\pi /6}^{\pi /3}1dx={\left[x\right]}_{\pi /6}^{\pi /3}=\frac{\pi }{6}\therefore I=\frac{\pi }{12}.$