Discussion Forum

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

Answer:

Explanation:

I = $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{1}{1+\sqrt{\cot x}}dx$

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx$ ...(i)

Then I=

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\sqrt{\sin (\frac{\pi}{2}-x)}}{\sqrt{\sin(\frac{\pi}{2}-x)}+\sqrt{\cos(\frac{\pi}{2}-x)}}dx$ 

I = $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\sqrt{\cos x)}}{\sqrt{\cos x)}+\sqrt{\sin x)}}dx$ ...(ii)

Adding (i) and (ii) we get

2I = $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\sqrt{\sin x}+\sqrt{\cos x)}}{\sqrt{\cos x)}+\sqrt{\sin x)}}dx$

$2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}1.dx =[x]_\frac{\pi}{6}^\frac{\pi}{3}$

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

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