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.
 A: $$\int_0^{\frac{\pi}2} \frac{\sqrt[3]{\tan x}}{(1+ \tan x)^2}\frac{dx}{\cos^2x}$$
$$=\int_0^\infty\frac{t^{1/3}}{(1+t)^2}dt=\int_0^\infty\frac s{(1+s^3)^2}3s^2ds,$$
and you certainly know how to integrate a rational fraction.
A: $u = \tan^\frac{1}{3}x,\\
3~\mathrm du = \tan^\frac{-2}{3}x \sec^2x ~\mathrm dx$
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.
A: 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,
$$I(\frac{1}{3}  ,1)=\Gamma(\frac{4}{3} )\Gamma(\frac{2}{3} )= \frac{1}{3}  \Gamma(\frac{1}{3} )\Gamma(\frac{2}{3} ) =\frac{2\pi}{3\sqrt 3} $$
A: There are branch-cut issues but I tried this in the given short time (assuming I am in the exam):
Let $I=\int_0^{\frac{\pi}{2}} \frac{\sqrt[3]{\tan(x)}}{(\cos(x) + \sin(x))^2}dx=\int_0^\infty\frac{z^{1/3}}{(1+z)^2}dz=\int_0^\infty\frac{2z^{5/3}}{(1+z^2)^2}dz$. Then by using the contour integration along the famous counter-clockwise closed semi-circle contour on the upper half plane and the residue theorem:
$$\int_{-\infty}^\infty\frac{2z^{5/3}}{(1+z^2)^2}dz=(e^{5\pi i/3}+1)I=2\pi iRes_{z=i}\frac{2z^{5/3}}{(1+z^2)^2}=2\pi i(-\frac13e^{\pi i/3})$$
Hence, $(e^{2\pi i/3}+e^{\pi i/3})=I=2\pi i/3\implies I=\frac{2\pi}{3\sqrt 3}.$
Note: Residue calculation
