$\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{2}{3}} x}{\sin ^{\frac{2}{3}} x+\cos ^{\frac{2}{3}} x} d x=$
$\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{2}{3}} x}{\sin ^{\frac{2}{3}} x+\cos ^{\frac{2}{3}} x} d x=$
$\frac{\pi}{4}$
$\frac{\pi}{8}$
$\frac{\pi}{2}$
$\pi$
Solution
$\begin{aligned} I &=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{2}{3}} x}{\sin ^{\frac{2}{3}} x+\cos ^{\frac{2}{3}} x} d x ...(1)\\ &=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{2}{3}}\left(\frac{\pi}{2}-x\right)}{\sin ^{\frac{2}{3}}\left(\frac{\pi}{2}-x\right)+\cos ^{\frac{2}{3}}\left(\frac{\pi}{2}-x\right)} d x \\ \therefore I &=\int_{0}^{2} \frac{\cos ^{\frac{2}{3}} x}{\sin ^{\frac{2}{3}} x+\cos ^{\frac{2}{3}} x} d x ...(2) \end{aligned}$
Adding equation (1) \& (2) we get
$\begin{aligned}
2 I &=\int_{0}^{2} 1 d x \Rightarrow 2 I=[x]_{0}^{\frac{\pi}{2}} \\
I &=\frac{1}{2}\left(\frac{\pi}{2}-0\right)=\frac{\pi}{4}
\end{aligned}$