Integrate $frac{theta sin theta}{1+cos^2 theta}$ with respect to $theta$

Alternatively, just do a simple integration by parts with $u=theta$ and $dv=frac{sin{theta} dtheta}{1+cos^2{theta}}$: $$int_0^pi frac{theta sin theta}{1+cos^2 theta} dtheta= -theta arctan{left(cos {theta}right)}bigg rvert_0^{pi}+int_0^{pi} arctan{left(cos {theta}right)} dtheta$$ For the second integral, notice that it is odd about $theta=frac{pi}{2}$ or if you don't see that then $theta mapsto theta-frac{pi}{2}$ $$=frac{pi^2}{4} + require{cancel} cancel{int_{-frac{pi}{2}}^{frac{pi}{2}} arctan{left(sin{theta}right)} d theta}$$ $$=boxed{frac{pi^2}{4}}$$

Edit: I got $arctan{left(cos{theta}right)}$ by substituting $u=cos{theta}$ for the $dv$ expression. The $sin{theta}$ cancels from the $du$ expression and it's just a straightforward $arctan{u}$ integral. As Barry Chipa said in the comments, the second integral is odd (substitute $xi=-theta$ to see this (remember that both $sin{theta}$ and $arctan{theta}$ are odd functions.

Another really cool way of getting the answer is from infinite series!!!!

Notice that

$${frac{1}{1+cos^2(x)}=sum_{n=0}^{infty}left(-cos^2(x)right)^n}$$

And so

$${int_{0}^{pi}frac{xsin(x)}{1+cos^2(x)}dx=int_{0}^{pi}xsin(x)sum_{n=0}^{infty}left(-cos^2(x)right)^ndx}$$

After interchanging a few things around, the integral becomes

$${=sum_{n=0}^{infty}(-1)^nint_{0}^{pi}xsin(x)cos^{2n}(x)dx}$$

If we use integration by parts on the inner integral, with ${dv=sin(x)cos^{2n}(x)dx}$ and ${u=x}$ you end up with

$${int_{0}^{pi}(-1)^nxsin(x)cos^{2n}(x)dx=(-1)^nleft(left(xfrac{-cos^{2n+1}(x)}{2n+1}right)_{x=0}^{x=pi} + frac{1}{2n+1}int_{0}^{pi}cos^{2n+1}(x)dxright)}$$

The rightmost integral will always be zero, and so we just end up with

$${=frac{(-1)^npi}{2n+1}}$$

Hence overall

$${int_{0}^{pi}frac{xsin(x)}{1+cos^2(x)}dx=pi sum_{n=0}^{infty}frac{(-1)^n}{2n+1}}$$

The infinite sum is just the Leibniz infinite series for ${frac{pi}{4}}$. So

$${=pifrac{pi}{4}=frac{pi^2}{4}}$$

Here's a trick which I use always with integrals involving trigonometric functions : $$int_alpha^beta varphi (xi) dxi=int_alpha^beta varphi (alpha +beta-xi) dxi$$ The proof is trivial and left for you as an exercise, lol!

Anyway, applying this technique to this integral :

Let $$I=int_0^pi frac{xsin x}{1+cos^2x}dx$$ We'll have after applying this formula : begin{align} I&=int_0^pi frac{(pi-x)sin (pi-x)}{1+cos^2(pi-x)} dx\ 2I&=int_0^pi frac{xsin x}{1+cos^2x} + frac{(pi-x)sin (pi-x)}{1+cos^2(pi-x)}dx\ I&=frac{1}2int_0^pi frac{x sin x+pi sin x-xsin x}{1+cos^2x}\ &=frac{1}2int_0^pi frac{pi sin x}{1+cos^2x}\ &=frac{pi}2int_0^pi frac{ sin x}{1+cos^2x} end{align} Now using the substitution you did earlier $$ u=cos x Leftrightarrow du=-sin x$$ So ; begin{align} I&=frac{pi}2int_1^{-1} frac{-du}{1+u^2}\ &=frac{pi}2int_{-1}^{1} frac{du}{1+u^2}\ &=frac{pi}2 arctan ubiggvert_{-1}^1\ &=frac{pi}2 bigg(frac{pi}4 +frac{pi}4bigg)\ &=frac{pi^2}{4} end{align} Hence as @PeterForeman said your integral is : $displaystyle frac{pi^2}{4}$

By the way, if you want the proof of the formula, all that you have to do is : $$xi=alpha +beta-u Leftrightarrow dxi=-du$$ Therefore;

$$int_beta^alpha varphi (alpha+beta-u) (-du)=int_alpha^beta varphi (alpha +beta-xi) dxi$$

$$I=int_0^pi frac{theta sin theta}{1+cos^2 theta} dthetatag 1$$ Using property of definite integral: $int_a^bf(x)dx=int_a^bf(a+b-x)dx$, $$I=int_0^pi frac{(pi-theta) sin theta}{1+cos^2 theta} dthetatag 2$$ Adding (1) and (2), $$2I=int_0^pi frac{pi sin theta}{1+cos^2 theta} dtheta$$ $$I=frac{pi}{2}int_0^{pi} frac{ sin theta dtheta}{1+cos^2 theta} $$ $$I=- piint_0^{pi/2} frac{ d(cos theta)}{1+cos^2 theta} $$ $$I=-pileft[tan^{-1}left(costhetaright)right]_0^{pi/2}$$ $$=frac{pi^2}{4}$$