Evaluating $\int _0^{\frac{\pi }{2}}\:\frac{\sqrt[3]{\sin^2\left(x\right)}}{\sqrt[3]{\sin^2\left(x\right)}+\sqrt[3]{\cos^2\left(x\right)}}dx$ [duplicate]

We have $$\begin{align} I=\int_{0}^{\pi/2}\frac{\sqrt[3]{\sin^{2}\left(x\right)}}{\sqrt[3]{\sin^{2}\left(x\right)}+\sqrt[3]{\cos^{2}\left(x\right)}}dx= & \int_{0}^{\pi/2}\frac{\sqrt[3]{\sin^{2}\left(x\right)}+\sqrt[3]{\cos^{2}\left(x\right)}-\sqrt[3]{\cos^{2}\left(x\right)}}{\sqrt[3]{\sin^{2}\left(x\right)}+\sqrt[3]{\cos^{2}\left(x\right)}}dx \\ = & \frac{\pi}{2}-\int_{0}^{\pi/2}\frac{\sqrt[3]{\cos^{2}\left(x\right)}}{\sqrt[3]{\sin^{2}\left(x\right)}+\sqrt[3]{\cos^{2}\left(x\right)}}dx. \end{align}$$ And now since holds $$\int_{a}^{b}f\left(x\right)dx=\int_{a}^{b}f\left(a+b-x\right)dx $$ we have $$I=\frac{\pi}{2}-\int_{0}^{\pi/2}\frac{\sqrt[3]{\cos^{2}\left(\frac{\pi}{2}-x\right)}}{\sqrt[3]{\sin^{2}\left(\frac{\pi}{2}-x\right)}+\sqrt[3]{\cos^{2}\left(\frac{\pi}{2}-x\right)}}dx=\frac{\pi}{2}-\int_{0}^{\pi/2}\frac{\sqrt[3]{\sin^{2}\left(x\right)}}{\sqrt[3]{\sin^{2}\left(x\right)}+\sqrt[3]{\cos^{2}\left(x\right)}}dx $$ hence $$\int_{0}^{\pi/2}\frac{\sqrt[3]{\sin^{2}\left(x\right)}}{\sqrt[3]{\sin^{2}\left(x\right)}+\sqrt[3]{\cos^{2}\left(x\right)}}dx=\frac{\pi}{4}. $$

Hint. $$ I = \int_{0}^{\pi/2}\frac{\sqrt[3]{\sin^{2}(x)}}{\sqrt[3]{\sin^{2}(x)}+\sqrt[3]{\cos^{2}(x)}} =\int_{0}^{\pi/2}\frac{\sqrt[3]{\cos^{2}(x)}}{\sqrt[3]{\sin^{2}(x)}+\sqrt[3]{\cos^{2}(x)}}$$ Now add both the integrals and see. We are basically using the following porperty of the definite integral $$\int_{0}^{a} f(x) \ dx = \int_{0}^{a} f(a-x) \ dx$$

See. we can write $\int f(x)=\int f(a+B-x)$ where a,b are the limits of integration. Now note that even with this the denominator remains the same as $cos(90-x)=sin(x)$ and also $sin(90-x)=cos(x)$ so let the integral be called I hence the one with $f(a+b-x)$ is also $I$. Adding these two integrals we get $2I=\int dx$ so $I=\frac{\pi}{4}$