# Evaluate $\int^{\frac{\pi}{3}}_{\frac{\pi}{6}} \frac{\sqrt[3]{\sin x}}{\sqrt[3]{\sin x} + \sqrt[3]{\cos x}} dx$.

Evaluate $$\int^{\frac{\pi}{3}}_{\frac{\pi}{6}} \frac{\sin^{1/3} x}{\sin^{1/3} x + \cos^{1/3} x} dx.$$

I tried doing by various methods but the answer is not coming.

• Let me edit!... Commented Sep 9, 2017 at 10:41
• Oh..thanks dear. Was trying to do but it was not fitting right. Commented Sep 9, 2017 at 10:43
• Use $\int_{a}^{b}\mathrm{f}\left(x\right)\mathrm{d}x = \int_{a}^{b}\mathrm{f}\left(a + b - x\right)\mathrm{d}x$. Commented Sep 19, 2017 at 18:34

Hint: $$I=\int^{\frac{\pi}{3}}_{\frac{\pi}{6}} \frac{\sin^{1/3} x}{\sin^{1/3} x + \cos^{1/3} x} dx$$ let $x=\pi/2-u$, this should give you $2I = \pi/6$.

• Why 2I since I1 and I2 are not the same as they differ by $$sin^(1/3) x$$ and another Integral has in its numerator $$cos^(1/3) x$$ Commented Sep 9, 2017 at 10:50
• I got it. Thanks. Commented Sep 9, 2017 at 10:59

Consider $$I=\int^{\frac{\pi}{3}}_{\frac{\pi}{6}} \frac{\sin^{1/3} x}{\sin^{1/3} x + \cos^{1/3} x} dx\quad \mbox{and}\quad J=\int^{\frac{\pi}{3}}_{\frac{\pi}{6}} \frac{\cos^{1/3} x}{\sin^{1/3} x + \cos^{1/3} x} dx.$$ Then we easily find that $I+J=\int^{\frac{\pi}{3}}_{\frac{\pi}{6}}1dx=\frac{\pi}{6}.$

Recalling that $\sin(\pi/2-x)=\cos(x)$ we have $$J=\int^{\frac{\pi}{3}}_{\frac{\pi}{6}} \frac{\sin^{1/3} (\pi/2-x)}{\cos^{1/3} (\pi/2-x) + \sin^{1/3} (\pi/2-x)} dx.$$ What do we obtain by letting $t=\pi/2-x$?

What is the value of $I$?

• Ok. Give me some time to calculate I - J . Thanks for the hint . Commented Sep 9, 2017 at 10:52
• @Sitanshu For $I-J$ consider the subtitution given in my edited answer. Commented Sep 9, 2017 at 10:55
• Thanks . Yes, J = I . That means 2I = π/6 . Thanks a lot. Commented Sep 9, 2017 at 10:59
• @Sitanshu Perfect. Well done. Commented Sep 9, 2017 at 11:00