Tricky trig integral

Question

I want to know how one should evaluate the following integral: $$\int_{0}^{\pi} \frac{x\cdot\text{sin}(x)}{1+\text{cos}^2(x)} dx$$

It doesn't lend itself to substitution or integration by parts; I also haven't been able to find any "cheap shots" using properties of odd and even functions. I tried using Feynman's trick with it, but it doesn't seem very helpful here (where to introduce the arbitrary variable other than in the power on cosine?).

I want to know how to solve this integral, but also more broadly, if there is any transferrable skill/trick I can learn from it. I am also interested in how one can discern which integrals are good candidates for Feynman's trick.

Whenever I see in the denominator an $x$ multiplied by something, e.g. $\int_a^b \frac{f(tx)}{x g(x)}dx$, I like to differentiate with respect to a parameter, say $t$, so that I could get rid of the $x$ and evaluate a possibly easier expression like $\int_a^b \frac{f'(xt)}{g(x)}dx$. — Accelerator, Jul 27 '23 at 21:03
Notice that if the $x$ wasn't there you would want to do $u=\cos x$. But the extra $x$ is a bit annoying. To get rid of it, you can do partial integration, which will give you a bit work to integrate $\frac{\sin x}{1+\cos^2x}$ but nothing too complex, that should give you an easier integral (I think). That's what I would try. — Vincent Batens, Jul 27 '23 at 21:05
@Accelerator I agree, but I don't see how that would be applicable here. $x$ is in the numerator right? — Vincent Batens, Jul 27 '23 at 21:07
@VincentBatens Yes, but at the end of the post, the OP was asking how one could discern what expressions could be worth differentiating under the integral. In fact, I could have changed my first comment to where I had $\int_a^b \frac{f(t h(x))}{h(x) g(x)}dx$ instead and differentiating with respect to $t$ could be a worthy trick. — Accelerator, Jul 27 '23 at 21:12
My approach does not work. It gives you an ugly integral that's left. — Vincent Batens, Jul 27 '23 at 21:20
Some of the methods shown here (same integral without $\sin$) may also be helpful. — user170231, Jul 27 '23 at 21:57
Does this answer your question? Integrating $\int^0_\pi \frac{x \sin x}{1+\cos^2 x}$ - found using an Approach0 search. Note another duplicate is Evaluate $\int^{\pi}_0\frac{x\sin(x)}{1+\cos^2(x)}dx$, and also ... — John Omielan, Jul 28 '23 at 10:02
(cont.) Why can we redefine the definition of a variable during substitution? Or let say assume it has two different values is very similar. — John Omielan, Jul 28 '23 at 10:04

score 12 · Accepted Answer · answered Jul 27 '23 at 21:19

Let $x = \pi-u$. Then

$\displaystyle I = \int_{0}^{\pi} \frac{x\cdot\text{sin}(x)}{1+\text{cos}^2(x)}\, dx = \int_{0}^{\pi} \frac{(\pi-u)\cdot\text{sin}(u)}{1+\text{cos}^2(u)}\,du = \int_{0}^{\pi} \frac{(\pi-x)\cdot\text{sin}(x)}{1+\text{cos}^2(x)}\,dx$

So that

$\displaystyle 2I = \int_{0}^{\pi} \frac{x\cdot\text{sin}(x)}{1+\text{cos}^2(x)}\, dx +\int_{0}^{\pi} \frac{(\pi-x)\cdot\text{sin}(x)}{1+\text{cos}^2(x)}\, dx = \pi \int_{0}^{\pi} \frac{\text{sin}(x)}{1+\text{cos}^2(x)}\, dx$

The last integral should be routine to calculate (e.g. let $t = \cos{x}$).

score 7 · Answer 2 · answered Jul 27 '23 at 23:07

Symmetry $$ \begin{align} \int_0^\pi\frac{x\sin(x)}{1+\cos^2(x)}\,\mathrm{d}x &=\int_0^\pi\frac{(\pi-x)\sin(x)}{1+\cos^2(x)}\,\mathrm{d}x\tag{1a}\\ &=\frac\pi2\int_0^\pi\frac{\sin(x)}{1+\cos^2(x)}\,\mathrm{d}x\tag{1b}\\ &=\left.-\frac\pi2\arctan(\cos(x))\,\right|_0^\pi\tag{1c}\\ &=\frac{\pi^2}4\tag{1d} \end{align} $$ Explanation:
$\text{(1a):}$ substitute $x\mapsto\pi-x$
$\text{(1b):}$ average both sides of $\text{(1a)}$
$\text{(1c):}$ $\frac{\mathrm{d}}{\mathrm{d}x}\arctan(\cos(x))=-\frac{\sin(x)}{1+\cos^2(x)}$
$\text{(1d):}$ evaluate

Integration by Parts $$ \begin{align} \int_0^\pi\frac{x\sin(x)}{1+\cos^2(x)}\,\mathrm{d}x &=-\int_0^\pi x\,\mathrm{d}\arctan(\cos(x))\tag{2a}\\ &=\frac{\pi^2}4+\int_0^\pi\arctan(\cos(x))\,\mathrm{d}x\tag{2b}\\ &=\frac{\pi^2}4-\int_0^\pi\arctan(\cos(x))\,\mathrm{d}x\tag{2c}\\ &=\frac{\pi^2}4\tag{2d} \end{align} $$ Explanation:
$\text{(2a):}$ $\frac{\mathrm{d}}{\mathrm{d}x}\arctan(\cos(x))=-\frac{\sin(x)}{1+\cos^2(x)}$
$\text{(2b):}$ integrate by parts
$\text{(2c):}$ substitute $x\mapsto\pi-x$
$\text{(2d):}$ average $\text{(2b)}$ and $\text{(2c)}$

score 5 · Answer 3 · answered Jul 27 '23 at 22:02

$$\begin{align*} & \int_0^\pi \frac{x \sin x}{1 + \cos^2x} \, dx \\ &= \int_{-\tfrac\pi2}^{\tfrac\pi2} \frac{\left(y+\frac\pi2\right) \cos y}{1+\sin^2y} \, dy & x=y+\frac\pi2 \\ &= \frac\pi2 \int_{-\tfrac\pi2}^{\tfrac\pi2} \frac{\cos y}{1+\sin^2y} \, dy & \text{parity}\\ &= \frac\pi2 \int_{-1}^1 \frac{dz}{1+z^2} & z=\sin y \end{align*}$$

Tricky trig integral

3 Answers3