Question

Evaluate: ${\int }_{\pi /6}^{\pi /3}\frac{dx}{1+\sqrt{\cot x}}.$

Answer 1

Let $I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{dx}{1+\sqrt{cotx}}={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\sqrt{sinx}}{\sqrt{sinx}+\sqrt{cosx}}dx$

$={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\sqrt{sin(\frac{\pi }{3}-\frac{\pi }{6}-x)}}{\sqrt{sin(\frac{\pi }{3}-\frac{\pi }{6}-x)}+\sqrt{cos(\frac{\pi }{3}-\frac{\pi }{6}-x)}}$

$I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt{sinx}}dx$

Adding we get, $2I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}dx=[x{]}_{\frac{\pi }{6}}^{\frac{\pi }{3}}=\frac{\pi }{3}-\frac{\pi }{6}=\frac{\pi }{6}$

Therefore, $I=\frac{\pi }{12}.$