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.
 A: This integral appear in the Jozsef Wildt International mathematical competition proposed by  Ovidui Furdui and Alina Sintamarian which I solved in the following way , couples of months back.
For all $n\geq 2$ we shall show that

$$I(n)=\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin^{2n-1}x+\cos^{2n-1}x}dx=\frac{\pi}{2n-1}\sum_{k=0}^{n-2}{n-2\choose k}\operatorname{csc}\left(\frac{(2\pi(n-k-1)}{2n-1}\right)$$.

Before we prove the above closed form we shall be using classical result.
Lemma: For all $n>1, m\in\mathbb {N}$following holds

$$ \int_0^{\infty}\frac{x^{m-1}}{1+x^n}dx=\frac{\pi}{n}\operatorname{csc}\left(\frac{m}{n}\pi\right)$$

Proof: We make subbing of $\frac{1}{1+x^n} = y$ and the integral takes the form of beta function, i.e; $$ \frac{1}{n}\int_0^{1}y^{1-\frac{m-n}{n}}(1-y)^{\frac{m}{n}-1}dy=\frac{1}{n}\Gamma\left(1-\frac{m}{n}\right)\Gamma\left(\frac{m}{n}\right)=\frac{\pi}{n}\operatorname{csc}\left(\frac{m}{n}\pi\right) $$
We evaluate the main integral $I(n)$ as follows $$\int_{0}^{\frac{\pi}{2}}\frac{\sin x\sec^{2n-1}x}{1+\tan^{2n-1} x}dx=\int_0^{\frac{\pi}{2}}\frac{\tan x (\sec^{2}x)^{n-2}\sec^2x}{1+\tan^{2n-1}x}dx$$ substitute $\tan x =u\implies \sec^{2}x dx=du$  and hence $$\int_0^{\infty}\frac{u(u^2+1)^{n-2} du}{1+u^{2n-1}} =\int_0^{\infty}\frac{u^2}{1+u^{2n-1}}\sum_{k=0}^{n-2}{n-2\choose k} u^{2(n-2-k)}du =\sum_{k=0}^{n-2}{n-2\choose k}\left(\int_0^{\infty}\frac{u^{2n-2k-4+1}}{1+u^{2n-1}} du\right) \underbrace{=}_{Lemma}\frac{\pi}{2n-1}\sum_{k=0}^{n-2}{n-2\choose k}\operatorname {csc}\left(\frac{2(n-k-1)\pi}{2n-1}\right)$$ we are done.
For $n=3$ we have a beautiful closed form for above integral

$$\int_0^{\frac{\pi}{2}}\frac{\sin x}{\sin ^{5} x+\cos^5 x}dx = \frac{2}{5} \sqrt{1+\frac{2}{\sqrt 5}}\pi\approx 1.729 $$

A: 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}
