Evaluating the following integral from $\pi/6$ to $\pi/2$

Question

I've been trying to evaluate the following integral:

$$I = \int_{\pi/6}^{\pi/2} \frac{\sin^{1/3} x}{\sin^{1/3} x + \cos^{1/3} x}\, dx$$

The integral is proper in the given interval and thus converging. I've tried the following method: $$I = \int_{\pi/6}^{\pi/2} \frac{\tan^{1/3} x}{\tan^{1/3} x + 1}\, dx$$ Substituting $\tan^{1/3} x + 1 = t$, evaluating, followed by $t-1= w$ the integral becomes $$I = \int_{\sqrt[\leftroot{-2}\uproot{2}6]{1/3}}^{\infty} \frac{3 w^3}{(1+w)(1 + w^6)}\, dw$$

Please suggest how to proceed further.

Answer 1

Now, use $$\frac{u^3}{(u+1)(u^6+1)}=\frac{2u^3-u^2-u+2}{3(u^4-u^2+1)}-\frac{u+1}{6(u^2+1)}-\frac{1}{2(u+1)}$$ and $$u^4-u^2+1=(u^2+1)^2-3u^2=(u^2-\sqrt3u+1)(u^2+\sqrt3u+1).$$

Answer 2

from $$\int \frac{\tan(x)^{1/3}}{\tan(x)^{1/3}+1}dx$$ we get with $$\tan(x)^{1/3}+1=t$$ : $$dt=\frac{1}{3}\tan(x)^{-2/3}(1+\tan^2(x))dx$$ where $$\tan(x)=(t-1)^3$$