How to find this integral $I=\int_{-\pi}^{+\pi}\frac{x\sin{x}\arctan{e^x}}{1+\cos^2{x}}dx$? Find the integral
$$I=\int_{-\pi}^{+\pi}\dfrac{x\sin{x}\arctan{e^x}}{1+\cos^2{x}}dx$$
My try: 
let
$$I=\int_{-\pi}^{0}\dfrac{x\sin{x}\arctan{e^x}}{1+\cos^2{x}}dx+\int_{0}^{+\pi}\dfrac{x\sin{x}\arctan{e^x}}{1+\cos^2{x}}dx=I_{1}+I_{2}$$
for $I_{2}$,we let $x=-t$
 A: Putting $x=-t,$
$$I_1=\int_{-\pi}^0\frac{x\sin x\arctan(e^x)}{1+\cos^2x}dx$$
$$=\int_{\pi}^0\frac{(-t)\sin(-t)\arctan(e^{-t})}{1+\cos^2(-t)}(-dt)$$
$$=-\int_{\pi}^0\frac{t\sin t\arctan(e^{-t})}{1+\cos^2t}dt$$
$$=\int_0^{\pi}\frac{t\sin t\arctan(e^{-t})}{1+\cos^2t}dt$$
Now, $\arctan \frac1x=\text{arccot} x=\frac\pi2-\arctan x$
$$\implies I_1=\int_0^{\pi}\frac{t\sin t\left(\frac\pi2-\arctan(e^t)\right)}{1+\cos^2t}dt $$
$$\implies I=I_1+I_2=\frac\pi2 \int_0^{\pi}\frac{t\sin t}{1+\cos^2t}dt $$
$$\text{Applying }\int_a^bf(x)dx=\int_a^bf(a+b-x)dx\text{ on }I \text{ with }a=0,b=\pi$$
$$ I=\frac\pi2\int_0^{\pi}\frac{t\sin t}{1+\cos^2t}dt=\frac\pi2\int_0^{\pi}\frac{(\pi-t)\sin(\pi- t)}{1+\cos^2(\pi-t)}dt=\frac\pi2\int_0^{\pi}\frac{(\pi-t)\sin t}{1+\cos^2t}dt$$
$$\implies I=\frac{\pi}2\left(\int_0^{\pi}\frac{\pi\sin t}{1+\cos^2t}dt-I\right)$$
Can you take it from here?
A: $$I=\int_{-\pi}^{\pi}\frac{x\sin x\arctan e^x}{1+\cos^2x}dx$$
$$=\int_{-\pi}^{\pi}\frac{x\sin x\arctan e^{-x}}{1+\cos^2x}dx$$
$$=\frac{1}{2}\int_{-\pi}^{\pi}\frac{x\sin x(\arctan e^x+\arctan e^{-x})}{1+\cos^2x}dx$$
$$=\frac{\pi}{4}\int_{-\pi}^{\pi}\frac{x\sin x}{1+\cos^2x}dx$$
$$=\frac{\pi}{2}\int_{0}^{\pi}\frac{x\sin x}{1+\cos^2x}dx$$
$$=\frac{\pi^2}{2^2}\int_{0}^{\pi}\frac{\sin x}{1+\cos^2x}dx$$
$$=\frac{\pi^2}{2^2}(-\arctan(\cos x))|_0^\pi=\frac{\pi^3}{8}$$
