Question

Using properties of definite integrals, evaluate :

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

Answer 1

$I=\underset{\pi /6}{\overset{\pi /3}{\int }}\frac{1}{1+\sqrt{\cot x}}dx$

$I=\underset{\pi /6}{\overset{\pi /3}{\int }}\frac{1}{1+\frac{\sqrt{\cos x}}{\sqrt{\sin x}}}dx$

$I=\underset{\pi /6}{\overset{\pi /3}{\int }}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$ .....(1)

We know,

$\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$

Using this property in equation 1, we get,

$I=\underset{\pi /6}{\overset{\pi /3}{\int }}\frac{\sqrt{\sin (\frac{\pi }{6}+\frac{\pi }{3}-x)}}{\sqrt{\sin (\frac{\pi }{6}+\frac{\pi }{3}-x)}+\sqrt{\cos (\frac{\pi }{6}+\frac{\pi }{3}-x)}}dx$

$I=\underset{\pi /6}{\overset{\pi /3}{\int }}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx$ ....(2)

On adding equation 1 and 2, we get,

$2I=\underset{\pi /6}{\overset{\pi /3}{\int }}dx$

$2I=\frac{\pi }{3}-\frac{\pi }{6}$

$2I=\frac{\pi }{6}$

$I=\frac{\pi }{12}$