Integrate $\int_{0}^{2\pi} {{x\sin^{100}x}\over \sin^{100}x+\cos^{100}x}dx$

Question

$$\int_{0}^{2\pi} {{x\sin^{100}x}\over \sin^{100}x+\cos^{100}x}\,dx $$ Does anyone have any idea? I'm stuck on it.
Thanks!

Answer 1

HINT:

First use $I=\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx$

so that $I+I=\int_a^b[f(x)+f(a+b-x)]\ dx\ \ \ \ (1)$

Then partition the ranges as $[0,\pi/2];[\pi/2,\pi];[\pi,3\pi/2];[3\pi/2,2\pi]$

In each four ranges apply $(1)$ again.

Generalization: For, $$J=\int_{m\cdot\pi/2}^{(m+1)\pi/2}\dfrac{\sin^nx}{\cos^nx+\sin^nx}dx$$

$\sin\left(\dfrac{m\pi}2+\dfrac{(m+1)\pi}2-x\right)=\sin\left(m\pi+\dfrac\pi2-x\right)=(-1)^m\cos x$

Similarly, $\cos\left(\dfrac{m\pi}2+\dfrac{(m+1)\pi}2-x\right)=(-1)^m\sin x,$

If $g(x)=\dfrac{\sin^nx}{\cos^nx+\sin^nx},$

$g\left(\dfrac{m\pi}2+\dfrac{(m+1)\pi}2-x\right)=\cdots=\dfrac{\cos^nx}{\sin^nx+\cos^nx}$

$$\implies J+J=\int_{m\cdot\pi/2}^{(m+1)\pi/2}\dfrac{\sin^nx+\cos^nx}{\cos^nx+\sin^nx}dx=\cdots=\dfrac{(m+1)\pi}2-\dfrac{m\pi}2=?$$

Answer 2

Hint. By the change of variable $$ x \to 2\pi-x, $$ one gets $$ \int_{0}^{2\pi} {{x\sin^{100}x}\over \sin^{100}x+\cos^{100}x}\:dx =\int_{0}^{2\pi} {{(2\pi-x)\sin^{100}x}\over \sin^{100}x+\cos^{100}x}\:dx $$ then one may obtain a standard integral.

Answer 3

After the reduction (by symmetry, $x\mapsto 2\pi-x$) to $$\pi\int_{0}^{2\pi}\frac{\sin^{100}(x)}{\sin^{100}(x)+\cos^{100}(x)}\,dx=2\pi\int_{0}^{\pi}\frac{\sin^{100}(x)}{\sin^{100}(x)+\cos^{100}(x)}\,dx$$ we may still apply a symmetry argument ($x\mapsto \frac{\pi}{2}-x$) to turn the last integral into $$ 2\pi \int_{0}^{\pi/2}\frac{\sin^{100}(x)+\cos^{100}(x)}{\sin^{100}(x)+\cos^{100}(x)}\,dx = \color{red}{\large\pi^2}.$$ You may replace $100$ with any positive even integer and the answer stays the same.