About an integral of the MIT Integration Bee Finals (2023)

integrationtrigonometry

I would like to solve the first problem of the MIT Integration Bee Finals, which is the following integral :

$$\int_0^{\frac{\pi}{2}} \frac{\sqrt[3]{\tan(x)}}{(\cos(x) + \sin(x))^2}dx$$

I tried substitution $u=\tan(x)$, King Property, but nothing leads me to the solution which is apparently $\frac{2\sqrt{3}}{9} \pi$.

If anybody knows how to solve it I would be grateful.

Best Answer

$$\int_0^{\frac{\pi}2} \frac{\sqrt[3]{\tan x}}{(1+ \tan x)^2}\frac{dx}{\cos^2x}$$ $$=\int_0^\infty\frac{t^{1/3}}{(1+t)^2}dt=\int_0^\infty\frac s{(1+s^3)^2}3s^2ds,$$ and you certainly know how to integrate a rational fraction.

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