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.

I want to generalize the integral as
$$
I(a,n)=\int_0^{\frac{\pi}{2}} \frac{\tan^a x}{(\sin x+\cos x)^{2 n}}=\int_0^{\frac{\pi}{2}} \frac{\tan^a x}{(1+\sin 2 x)^n} d x
$$
where $0<a<1$ and $n\in N$.
Letting $t=\tan x$ gives
$$
I(a,n)=\int_0^{\infty} \frac{t^{a}}{\left(1+\frac{2 t}{1+t^2}\right)^n} \cdot \frac{d t}{1+t^2}=\int_0^{\infty} \frac{t^{a}\left(1+t^2\right)^{n-1}}{(t+1)^{2 n}} d t
$$
Using Binomial expansion and beta function, we have
$$\begin{aligned}
I(a,n)&=\sum_{k=0}^{n-1}\left(\begin{array}{c}
n-1 \\
k
\end{array}\right) \int_0^{\infty} \frac{t^{2 k+a}}{(t+1)^{2 n}} d t\\&=\sum_{k=0}^{n-1}\left(\begin{array}{l}
n \\
k
\end{array}\right) B\left(2 k+a+1,2n-2k-a-1  \right)\\&= \sum_{k=0}^{n-1}\left(\begin{array}{c}
n-1 \\
k
\end{array}\right) \frac{\Gamma\left(2 k+a+1\right) \Gamma\left(2 n-2 k—a-1\right)}{(2 n-1)!}\end{aligned}
$$
In particular,
$$I(\frac{1}{3}  ,1)=\Gamma(\frac{4}{3} )\Gamma(\frac{2}{3} )= \frac{1}{3}  \Gamma(\frac{1}{3} )\Gamma(\frac{2}{3} ) =\frac{2\pi}{3\sqrt 3} $$
