Question

The value of $\frac{36}{\pi }{\int }_{\pi /6}^{\pi /3}\frac{dx}{1+\sqrt{\cot x}}$ is

Answer 1

$I=\frac{36}{\pi }{\int }_{\pi /6}^{\pi /3}\frac{dx}{1+\sqrt{\cot x}}=\frac{36}{\pi }{\int }_{\pi /6}^{\pi /3}\frac{\sqrt{\sin x}dx}{\sqrt{\sin x}+\sqrt{\cos x}}$ ...(1)
Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
$I=\frac{36}{\pi }{\int }_{\pi /6}^{\pi /3}\frac{\sqrt{\sin (\frac{\pi }{2}-x)}dx}{\sqrt{\sin (\frac{\pi }{2}-x)}+\sqrt{\cos (\frac{\pi }{2}-x)}}=\frac{36}{\pi }{\int }_{\pi /6}^{\pi /3}\frac{\sqrt{\cos x}dx}{\sqrt{\cos x}+\sqrt{\sin x}}$ ...(2)
$2I=\frac{36}{\pi }{\int }_{\pi /6}^{\pi /3}dx=\frac{36}{\pi }{\left[x\right]}_{\pi /6}^{\pi /3}=6\Rightarrow I=3$