On the way of solving Integral $\int_{0}^{2\pi}\frac{x\sin^{2n}x}{\sin^{2n}x+\cos^{2n}x}dx$. Question

For $n > 0$, Find 
  $$\int_{0}^{2\pi}\frac{x\sin^{2n}x}{\sin^{2n}x+\cos^{2n}x}dx$$

My Approach
Let
 $I_{n}$  =$$\int_{0}^{2\pi}\frac{x \sin^{2n}x}{\sin^{2n}x+\cos^{2n}x}dx$$
I$_{n}$=$\int_{0}^{2\pi}$$\frac{\left(2\pi-x\right)\sin^{2n}x}{\sin^{2n}x+\cos^{2n}x}dx$$\Longrightarrow$2I$_{n}=$$\int_{0}^{2\pi}$$\frac{2\pi \sin^{2n}x}{\sin^{2n}x+\cos^{2n}x}dx$
$\Longrightarrow$I$_{n}=\int_{0}^{2\pi}\frac{\pi \sin^{2n}x}{\sin^{2n}x+\cos^{2n}x}dx$
I am stuck here!

I will assume $n \in \mathbb{N}$. If in the integral
$$I_n = \int^{2\pi}_0 \frac{x \sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx,$$
we set $x \mapsto 2\pi - x$ it can be readily seen that
$$I_n = \pi \int^{2\pi}_0 \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx.$$
Note the integrand
$$f(x) = \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x},$$
is both an even and periodic function with a fundamental period of $\pi$. If $f$ is a continuous bounded function with period $\mathfrak{a}$, then
$$\int^{b + \mathfrak{a}}_b f(x) \, dx = \int^\mathfrak{a}_0 f(x) \, dx,$$
where $b \in \mathbb{R}$.
Using this result repeatedly on our integral we have
\begin{align*}
I_n &= \pi \int^{-\pi + 2\pi}_{-\pi} \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx \qquad \text{(periodic with period $2\pi$)}\\
&= \pi \int^\pi_{-\pi} \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx\\
&= 2\pi \int^\pi_0 \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx \qquad \text{(since it is even)}\\
&= 2\pi \int^{-\pi/2 + \pi}_{-\pi/2} \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx \qquad \text{(periodic with period $\pi$)}\\
&= 4 \pi \int^{\pi/2}_0 \frac{\sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx \qquad \text{(since it is even)}.
\end{align*}
Now setting $x \mapsto \dfrac{\pi}{2} - x$ gives
$$I_n = 4 \pi \int^{\pi/2}_0 \frac{\cos^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx.$$
Thus
$$I_n + I_n = 4\pi \int^{\pi/2}_0 \frac{\sin^{2n} x + \cos^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx = 4\pi \int^{\pi/2}_{0} \, dx,$$
or
$$I_n = \int^{2\pi}_0 \frac{x \sin^{2n} x}{\sin^{2n} x + \cos^{2n} x} \, dx = \pi^2.$$
