What is the simplest technique to evaluate the following definite triple integral? Consider the following definite triple integral:
$$ \int_0^\pi \int_0^\pi \int_0^\pi \frac{x\sin x \cos^4y \sin^3z}{1 + \cos^2x} ~dx~dy~dz $$
According to Wolfram Alpha, this evaluates to $\frac{\pi^3}{8}$, but I have no idea how to obtain this result. The indefinite integral $$ \int \frac{x \sin x}{1 + \cos^2 x}~dx $$ appears to not be expressible in terms of elementary functions. Thus, I am at a loss as to what sort of techniques might be used to evaluate this integral. For context, this is from a past year's vector calculus preliminary exam at my graduate school, so while I'm sure there are some advanced integration techniques that can be used here, I'm particularly interested in what elementary techniques might be used to evaluate the integral, as I don't think something like, for instance, residue techniques would be considered pre-requisite knowledge for taking this exam. 
 A: First off, note that the integrals w.r.t. $y$ and $z$ are quite trivial to evaluate. Then, consider $x\mapsto\pi-x$, since trig functions are symmetric about $\pi/2$.
$$I=\int_0^\pi\frac{x\sin(x)}{1+\cos^2(x)}~\mathrm dx=\int_0^\pi\frac{(\pi-x)\sin(x)}{1+\cos^2(x)}~\mathrm dx$$
Add these together and apply $\cos(x)\mapsto x$.
$$\begin{align}\frac2\pi I&=\int_0^\pi\frac{\sin(x)}{1+\cos^2(x)}~\mathrm dx\\&=\int_{-1}^1\frac1{1+x^2}~\mathrm dx\\&=\arctan(1)-\arctan(-1)\\&=\frac\pi2\end{align}\\\implies I=\frac{\pi^2}4$$
A: Let $I = \int_0^{\pi} xf(\sin x) \, \mathrm{d}x$ then the sub $x \mapsto \pi -x$ gives $I=\int_{0}^{\pi} (\pi - x)f(\sin x) \, \mathrm{d}x$ so that $$I = \int_0^{\pi} \pi f(\sin x) \, \mathrm{d}x -I \iff I = \frac{\pi}{2}\int_0^{\pi} f(\sin x) \, \mathrm{d}x.$$
Note that $\cos^2 x=1-\sin^2 x$, so your integrand really is of the form$f(\sin x).$ Specializing to your case gives $$\int_0^{\pi} \frac{x\sin x}{1 + \cos^2 x} \, \mathrm{d}x = \frac{\pi}{2}\int_0^{\pi} \frac{\sin x}{1 + \cos^2 x} \, \mathrm{d}x.$$
This final integral succumbs immediately to $x \mapsto \cos x$, giving $$I = \frac{\pi}{2}\big[\arctan x\big]_{-1}^{1} = \frac{\pi^2}{4}$$
