$$\int_{\pi/6}^{\pi/2} \frac{\sin^{1/3} (x)}{\sin^{1/3} (x)+\cos^{1/3} (x)}$$

I tried using the substitution $t^3=\tan x$. Which gives me


How should I proceed?

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

I will put you on the path:

\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}

  • $\begingroup$ Sorry I didn't finish it, I was falling asleep as I was typing. Are you ok from here? or would you like some further assistance to move forward? Will type up if so :-) $\endgroup$ – user150203 Dec 20 '18 at 0:36

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.

  • $\begingroup$ I agree with you, Maxima gives the indefinite integral in terms of roots of $t^4-t^2+1=0$ $\endgroup$ – Awe Kumar Jha Dec 19 '18 at 12:37
  • $\begingroup$ Is there a simpler way to calculate the definite integral. I tried using this method and it is too long from an exam point of view. $\endgroup$ – Piyush Divyanakar Dec 19 '18 at 15:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.