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.

  • $\begingroup$ Well, the problem more or less reduces down to the integral w.r.t. $x$, since everything else is easy to integrate. $\endgroup$ – Simply Beautiful Art Aug 19 '17 at 22:23
  • $\begingroup$ Right, and according to Wolfram Alpha, that part of the integral evaluates to $\frac{\pi^2}{4}$. So I'd accept a solution of that as a solution to my problem. I don't have an issue with the rest of the integral. $\endgroup$ – Nathan BeDell Aug 19 '17 at 22:28

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 $\cos(x)\mapsto x$.

$$\begin{align}\frac2\pi I&=\int_0^\pi\frac{\sin(x)}{1+\cos^2(x)}~\mathrm dx\\&=\int_{-1}^1\frac1{1+x^2}~\mathrm dx\\&=\arctan(1)-\arctan(-1)\\&=\frac\pi2\end{align}\\\implies I=\frac{\pi^2}4$$

  • $\begingroup$ Totally bad at noticing and probably pointless, but I am in slight wonder as to why this was downvoted. $\endgroup$ – Simply Beautiful Art Aug 20 '17 at 23:59

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 $$I = \frac{\pi}{2}\big[\arctan x\big]_{-1}^{1} = \frac{\pi^2}{4}$$


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