How to evaluate $\int_0^{\pi/2} \frac{\sin x}{\sin^{2n+1}x +\cos^{2n+1}x} dx$? I have an exercise to evalute the following integral for all $n\geq 1 $
$$I(n)=\int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin^{2n+1} x+\cos^{2n+1} x}dx$$
I attempted to find the closed form for the integral above in the following manner, where I used the integral identity $\int_a^bf(x)=\int_a^b f(a+b-x)dx$.  $$I(\bar{n})=\int_0^{\frac{\pi}{2}}\frac{\cos x}{\cos^{2n-1} x+\sin ^{2n-1} x}dx$$ adding $I(n)$ and $I(\bar{n})$ its reduces to $$\frac{1}{2}\int_0^{\frac{\pi}{2}}\frac{\cos x +\sin x}{\cos^{2n+1}x +\sin^{2n+1}x}dx$$ using the algebraic identity  $a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^2+\cdots +b^{n-1})$ for odd integers $n$, I get $$\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{1}{\cos^{2n}x-\cos^{2n-1}\sin x+\cdots +\sin^{2n}x}dx $$
I'm now stuck here. How can I continue now?. Thanks in advance.

Utilize the decomposition
$$\frac{\sin x +\cos x}{\sin^{2n+1}x +\cos^{2n+1}x} 
=\frac{2^{n+1}}{2n+1}\sum_{k=1}^n\frac{(-1)^{k+1}\sin\frac{a_k}2 \cos^{n-1}a_k}{\csc a_k+\cot a_k \sin 2x}
$$
with $a_k=\frac{2\pi k}{2n+1}$ and integrate piecewise to arrive at
\begin{align}
\int_0^{\pi/2} \frac{\sin x}{\sin^{2n+1}x +\cos^{2n+1}x} dx
=\frac{2^{n+1}}{2n+1}\sum_{k=1}^n (-1)^{k+1}\ a_k \sin\frac{a_k}2 \cos^{n-1}a_k
\end{align}
The general result above produces
\begin{align}
\int_0^{\pi/2} \frac{\sin x}{\sin^{3}x +\cos^{3}x} dx 
&= \frac{2\pi}{3\sqrt3}\\
 \int_0^{\pi/2} \frac{\sin x}{\sin^{5}x +\cos^{5}x} dx 
&= \frac{2\pi}{5} \sqrt{1+\frac{2}{\sqrt 5}}\\
 \int_0^{\pi/2} \frac{\sin x}{\sin^{7}x +\cos^{7}x} dx 
&= \frac{4\pi}7\left(\sin\frac\pi7+\sin\frac{3\pi}7\right)\\
 \int_0^{\pi/2} \frac{\sin x}{\sin^{9}x +\cos^{9}x} dx 
&= \frac{4\pi}{9\sqrt3}\left(1+4\cos\frac{\pi}9\right)\\
 \int_0^{\pi/2} \frac{\sin x}{\sin^{11}x +\cos^{11}x} dx 
&= \ \cdots
\end{align}
