Evaluate $\int_{-\pi}^{\pi} {2x(1+\sin x)\over1+\cos^2x} \,\, \mathrm{d}x$ This is my question,
$$\int_{-\pi}^{\pi} {2x(1+\sin x)\over1+\cos^2x} \,\, \mathrm{d}x$$
(1) $\pi^2\over4$    
(2) $\pi^2$
(3) zero
(4) $\pi\over2$
I first broke the function into parts:
$\int_{-\pi}^{\pi} {2x\over1+\cos^2x}$+
$\int_{-\pi}^{\pi} {2x\sin x\over1+\cos^2x}$.
Here
$\int_{-\pi}^{\pi} {2x\over1+\cos^2x}$being an odd function becomes zero and$\int_{-\pi}^{\pi} {2x\sin x\over1+\cos^2x}$being an even function can be written as 4$\int_{0}^{\pi} {x\sin x\over1+\cos^2x}dx$ and now I used the property $\int_{0}^{a} {f(a-x)}dx$. Then,I'm stuck.  
Please help me through this.
Please explain any steps that you give.
Thanks in advance.
 A: I'm not an expert on definite integrals such as this, but it looks like you're on the right track.  I don't think you quite finished what your property was, but I think I see the direction it was headed in.  So from
$$4\int_0^\pi\frac{x\sin xdx}{1+\cos^2x}$$
substitute
$$u=\pi-x,du=-dx$$
$$4\int_0^\pi\frac{x\sin xdx}{1+\cos^2x}=-4\int_\pi^0\frac{(\pi-u)\sin(\pi-u)du}{1+\cos^2(\pi-u)}=$$
$$4\int_0^\pi\frac{(\pi-u)\sin udu}{1+(-\cos u)^2}$$
Finally, since the variable is just a placeholder, let's just replace $u$ with $x$ again to get
$$I=4\int_0^\pi\frac{x\sin xdx}{1+\cos^2x}=4\int_0^\pi\frac{(\pi-x)\sin xdx}{1+\cos^2x}$$
Summing these gives
$$2I=4\int_0^\pi\frac{\pi\sin xdx}{1+\cos^2x}$$
$$u=\cos x,du=-\sin xdx$$
$$I=-2\pi\int_1^{-1}\frac{du}{1+u^2}=2\pi(\tan^{-1}u]_{-1}^1)=2\pi(\frac{\pi}4+\frac\pi4)=\pi^2$$
So assuming I've done everything right, I arrive at answer 2.
A: Using symmetry we get
$$J:=\int_{-\pi}^\pi{2x(1+\sin x)\over 1+\cos^2 x}\ dx=4\int_0^\pi x\>{\sin x\over 1+\cos^2 x}\ dx\ .$$
Partial integration then gives
$$J=4\left(-x\arctan(\cos x)\biggr|_0^\pi +\int_0^\pi\arctan(\cos x)\ dx\right)\ .$$
As $x\mapsto \cos x$ is odd with respect to $x={\pi\over2}$ the last integral vanishes, so that we are left with
$$J=-4\pi\arctan(\cos\pi)=-4\pi\bigl(-{\pi\over4}\bigr)=\pi^2\ .$$
A: Using $\displaystyle\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx\ \  \ \ (1)$
If $\displaystyle f(x)=\frac{2x(1+\sin x)}{1+\cos^2x}, f\left(-\pi+\pi-x\right)=f(-x)=\frac{-2x(1-\sin x)}{1+\cos^2(-x)}=\frac{2x\sin x-2x}{1+\cos^2x}$
$$I=\int_{-\pi}^{\pi}\frac{2x(1+\sin x)}{1+\cos^2x}\ dx=\int_{-\pi}^{\pi}\frac{2x\sin x-2x}{1+\cos^2x}\ dx$$
$$\implies I+I=4\int_{-\pi}^{\pi}\frac{x\sin x}{1+\cos^2x}\ dx$$
Again if $\displaystyle g(x)=\frac{x\sin x}{1+\cos^2x}, g(-x)=\frac{(-x)\sin(-x)}{1+\cos^2(-x)}=\frac{x\sin x}{1+\cos^2x}$ i.e., $g(x)$ is even
$$\text{ As for even }g(x),\int_{-a}^ag(x)dx=2\int_0^ag(x)dx,$$
$$\int_{-\pi}^{\pi}\frac{x\sin x}{1+\cos^2x}\ dx=2\int_0^{\pi}\frac{x\sin x}{1+\cos^2x}\ dx$$
Again, using $(1),$
$$J=\int_0^{\pi}\frac{x\sin x}{1+\cos^2x}\ dx=\int_0^{\pi}\frac{(\pi-x)\sin(\pi-x)}{1+\cos^2(\pi-x)}\ dx=\int_0^{\pi}\frac{(\pi-x)\sin x}{1+\cos^2x}\ dx$$
$$\implies J+J=\pi\int_0^{\pi}\frac{\sin x}{1+\cos^2x}\ dx$$
Set $\displaystyle \cos x=u$
