# How to evaluate $\int_{\pi/6}^{\pi/3}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}$?

$$\int_{\pi/6}^{\pi/3} \frac {\sqrt{\sin x}} {\sqrt{\sin x} + \sqrt{\cos x}}$$

Hint use $f (x)=f (a+b-x )$ where a,b are lower ,upper limits . Then everything is straightforward
Using the hint given by Archis Welankar: $$\int_a^b f(x) dx=\int_a^bf(a+b-x)dx;\\ \int_{\pi/6}^{\pi/3} \frac {\sqrt{\sin x}} {\sqrt{\sin x} + \sqrt{\cos x}}dx=\int_{\pi/6}^{\pi/3} \frac {\sqrt{\sin (\pi/2-x)}} {\sqrt{\sin (\pi/2-x)} + \sqrt{\cos (\pi/2-x)}}dx \Rightarrow \\ \underbrace{\int_{\pi/6}^{\pi/3} \frac {\sqrt{\sin x}} {\sqrt{\sin x} + \sqrt{\cos x}}dx}_{A}=\underbrace{\int_{\pi/6}^{\pi/3} \frac {\sqrt{\cos x}} {\sqrt{\cos x} + \sqrt{\sin x}}dx}_{B} \Rightarrow \\ A-B=0.$$ Also: $$\int_{\pi/6}^{\pi/3} \frac {\sqrt{\sin x}} {\sqrt{\sin x} + \sqrt{\cos x}}dx=\int_{\pi/6}^{\pi/3} \frac {\sqrt{\sin x}+\sqrt{\cos x}-\sqrt{\cos x}} {\sqrt{\sin x} + \sqrt{\cos x}}dx=\\ \int_{\pi/6}^{\pi/3} 1-\frac {\sqrt{\cos x}} {\sqrt{\cos x} + \sqrt{\sin x}}dx \Rightarrow \\ A+B=\frac{\pi}{6}.$$ Hence: $$A=\frac{\pi}{12}.$$