Find a closed form for the following integral:
$$\int _{\pi/6} ^{\pi /3} \frac {\sqrt[3]{\sin x}}{\sqrt [3]{\sin x} + \sqrt[3]{\cos x}}dx$$
I think the answer is found out using some properties. So if it was $\frac\pi2$ the question would be easy.
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1Related : http://math.stackexchange.com/questions/578957/definite-integral-int-24-frac-sqrt-log9-x-sqrt-log9-x-sqrt-log3 and http://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3xdx – lab bhattacharjee Feb 14 '16 at 12:35
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An integral of the form $$I=\int_a^b \frac{f\left(x\right)}{f\left(x\right)+f\left(a+b-x\right)}dx$$ satisfies, by the substitution $y=a+b-x$,$$I=\int_a^b \frac{f\left(a+b-y\right)}{f\left(y\right)+f\left(a+b-y\right)}dy.$$Averaging gives $$I=\int_a^b \frac{dx}{2} = \frac{b-a}{2}.$$In the case at hand $$a=\frac{\pi}{6},\,b=\frac{\pi}{3},\,a+b=\frac{\pi}{2},\,f\left(x\right)= \sqrt[3]{\sin x},\,f\left(a+b-x\right)= \sqrt[3]{\cos x},\, \\ I=\frac{1}{2}\left(\frac{\pi}{3}-\frac{\pi}{6}\right)=\frac{\pi}{12}.$$
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