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

$u = \tan^\frac{1}{3}x,\\
3~\mathrm du = \tan^\frac{-2}{3}x \sec^2x ~\mathrm dx$
Shortens it down to -
$$\int \frac{3u^3~\mathrm du}{(1+u^3)^2}$$
Split it into
$\displaystyle\int \frac{u \cdot 3u^2~\mathrm du}{(1+u^3)^2}$ and perform by parts.
