Integral$\int_{\pi/6}^{\pi/2} \frac{\sin^{1/3} (x)}{\sin^{1/3} (x)+\cos^{1/3} (x)}$ $$\int_{\pi/6}^{\pi/2} \frac{\sin^{1/3} (x)}{\sin^{1/3} (x)+\cos^{1/3} (x)}$$
I tried using the substitution $t^3=\tan x$. Which gives me 
$$\int\frac{3t^3}{(t+1)(t^6+1)}$$
How should I proceed?

As Lord Shark the Unknown commented, the first thing to do is partial fraction decomposition
Since $t^6+1=(t^2+1)(t^4-t^2+1)$, then
$$\frac{3t^3}{(t+1)(t^6+1)}=-\frac{3}{2 (t+1)}-\frac{t+1}{2 \left(t^2+1\right)}+\frac{2 t^3-t^2-t+2}{t^4-t^2+1}$$I suppose that the last term will impose to work with the roots of unity.
