Can I get some help with this integral? $\int_{0}^{2\pi}\dfrac{x\sin^{100}(x)}{\cos^{100}(x)+\sin^{100}(x)}dx$ (I'm still trying to understand this concept) I tried to separate the integral, as $cos(x)\geq sin(x)$ in $[0,\frac{\pi}{4}]$ and $[\frac{5\pi}{4},2\pi]$, therefore $\sin^{100}(x)$ goes to $0$ faster than $\cos^{100}(x)$ does. 
Like this, $\dfrac{sin^{100}(x)}{\cos^{100}(x)+\sin^{100}(x)} \to{0}$ so the integral will be $0$.
On the other hand, in $[\frac{\pi}{4},\frac{5\pi}{4}]$ this $\dfrac{sin^{100}(x)}{\cos^{100}(x)+\sin^{100}(x)}\to{1}$ so the integral will be $\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}}x dx$. Like this, I will get $\frac{25\pi^{2}}{32} - \frac{\pi^{2}}{32} = \frac{24\pi^{2}}{32} = \frac{3\pi^{2}}{4}$ which is not the answer...
 A: Let $$I=\int_{0}^{2\pi}\dfrac{x\sin^{100}(x)}{\cos^{100}(x)+\sin^{100}(x)}dx$$
Now, make the substitution $x=2\pi-u$ to get $$\begin{align}I&=-\int_{2\pi}^{0}\dfrac{(2\pi-u)\sin^{100}(2\pi-u)}{\cos^{100}(2\pi-u)+\sin^{100}(2\pi-u)}du\\&=\int_{0}^{2\pi}\dfrac{(2\pi-u)\sin^{100}(u)}{\cos^{100}(u)+\sin^{100}(u)}du\\&=-I+\int_{0}^{2\pi}\dfrac{2\pi\sin^{100}(u)}{\cos^{100}(u)+\sin^{100}(u)}du\\\implies I&=\int_{0}^{2\pi}\dfrac{\pi\sin^{100}(u)}{\cos^{100}(u)+\sin^{100}(u)}du\\&=\int_{0}^{\pi}\dfrac{2\pi\sin^{100}(u)}{\cos^{100}(u)+\sin^{100}(u)}du\\&=\int_{0}^{\frac{\pi}{2}}\dfrac{4\pi\sin^{100}(u)}{\cos^{100}(u)+\sin^{100}(u)}du\end{align}$$
Where we made use of the periodicity. Now, use the fact that $\sin(\frac{\pi}{2}-x)=\cos(x)$ and $\cos(\frac{\pi}{2}-x)=\sin(x)$ and use the substitution $u=\frac{\pi}{2}-t$ to get $$I=-\int_{\frac{\pi}{2}}^{0}\dfrac{4\pi\sin^{100}(\frac{\pi}{2}-t)}{\cos^{100}(\frac{\pi}{2}-t)+\sin^{100}(\frac{\pi}{2}-t)}dt=\int_{0}^{\frac{\pi}{2}}\dfrac{4\pi\cos^{100}(t)}{\cos^{100}(t)+\sin^{100}(t)}dt$$
Now, add these two $I$'s together: $$I+I=\int_{0}^{\frac{\pi}{2}}\dfrac{4\pi\cos^{100}(t)+4\pi\sin^{100}(t)}{\cos^{100}(t)+\sin^{100}(t)}dt=4\pi\cdot\frac{\pi}{2}=2\pi^2\implies \boxed{I=\pi^2}$$
A: Let the integral be $I$ Using $f (x)=f (2\pi-x)$ and remembering $\sin (2\pi-x)=-\cos (x) $ we see that the cos cancel from numerator and denominator. Thus $I=\int _0 ^{2\pi} (\pi-\frac {x}{2}) =\pi ^2$
A: Here are some hints as to how to get the exactly answer $\pi^2$


*

*Do the substitution $u=2\pi-x$. This has the effect of removing the spare $x$ term from the integrand. You get $2I=2\pi\int_0^{2\pi}...$

*Consider symmetry and do $4\times \int_0^{\frac{\pi}{2}}...$

*Finally do another substitution $x=\frac{\pi}{2}-\theta$ and consider $$I+I=4\pi\int_0^{\frac{\pi}{2}}1d\theta$$
I hope this helps and you would like to try it for yourself
