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

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

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: $$\int_a^b f(x) dx=\int_a^bf(a+b-x)dx;\\ \int_{\pi/6}^{\pi/3} \frac {\sqrt[3]{\sin x}} {\sqrt[3]{\sin x} + \sqrt[3]{\cos x}}dx=\int_{\pi/6}^{\pi/3} \frac {\sqrt[3]{\sin (\pi/2-x)}} {\sqrt[3]{\sin (\pi/2-x)} + \sqrt[3]{\cos (\pi/2-x)}}dx \Rightarrow \\ \underbrace{\int_{\pi/6}^{\pi/3} \frac {\sqrt[3]{\sin x}} {\sqrt[3]{\sin x} + \sqrt[3]{\cos x}}dx}_{A}=\underbrace{\int_{\pi/6}^{\pi/3} \frac {\sqrt[3]{\cos x}} {\sqrt[3]{\cos x} + \sqrt[3]{\sin x}}dx}_{B} \Rightarrow \\ A-B=0.$$ Also: $$\int_{\pi/6}^{\pi/3} \frac {\sqrt[3]{\sin x}} {\sqrt[3]{\sin x} + \sqrt[3]{\cos x}}dx=\int_{\pi/6}^{\pi/3} \frac {\sqrt[3]{\sin x}+\sqrt[3]{\cos x}-\sqrt[3]{\cos x}} {\sqrt[3]{\sin x} + \sqrt[3]{\cos x}}dx=\\ \int_{\pi/6}^{\pi/3} 1-\frac {\sqrt[3]{\cos x}} {\sqrt[3]{\cos x} + \sqrt[3]{\sin x}}dx \Rightarrow \\ A+B=\frac{\pi}{6}.$$ Hence: $$A=\frac{\pi}{12}.$$