# 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?

• Partial fractions? By the way, you seem to have lost your limits. – Lord Shark the Unknown Dec 19 '18 at 5:42
• Typing on phone, so skipped it. Is a better way to type mathjax on phone – Piyush Divyanakar Dec 19 '18 at 5:43
• @Piyush Divyanakar, you should try with lower limit $\frac {π}{4}$ – Awe Kumar Jha Dec 19 '18 at 12:39

\begin{align} \frac{3x^3}{\left(x + 1\right)\left(x^6 + 1\right)} &= \frac{3x^3}{\left(x + 1\right)\left(x^2 + 1\right)\left(x^2 + \sqrt{3}x + 1\right)\left(x^2 - \sqrt{3}x + 1\right)} \\ &= -\frac{3}{2}\frac{1}{x + 1} - \frac{1}{2}\left[\frac{1}{x^2 + 1} +\frac{x}{x^2 + 1}\right] + \frac{1}{2\left(2 + \sqrt{3}\right)}\frac{x + 1}{x^2 - \sqrt{3}x + 1} \\ &\qquad + \frac{2 +\sqrt{3}}{2}\frac{x + 1}{x^2 + \sqrt{3}x + 1} \end{align}
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.
• I agree with you, Maxima gives the indefinite integral in terms of roots of $t^4-t^2+1=0$ – Awe Kumar Jha Dec 19 '18 at 12:37