Evaluate $\int_0^{\pi} \frac{x \sin x}{1+\cos ^2 x}dx$

Question

Find $$\int_0^{\pi} \frac{x \sin x}{1+\cos ^2 x}dx$$ by hand. I'm aware Mathematics gives $\frac{\pi^2}{4}$, but I need to learn this without the aid of Mathematica. I tried using substitutions like $u= \tan \frac{x}{2}$ and some trig identities, but I still can't work this out.

Answer 1

Let $$f(x)=\frac{\sin x}{{1+\cos^2x}}=\frac{\sin x}{{2-\sin^2x}}.$$ Note that $$f(π-t)=f(t).$$

Using $x=\pi-t$, $$\begin{align}I&=\int_{0}^{\pi} xf(x) dx\\&=\int_{0}^{\pi} (π-t) f(π-t) (-dt) \\&=\int_{0}^{\pi} (π-t) f(t) dt \\&=π\int_{0}^{\pi} f(t) dt-\int_{0}^{\pi}tf(t) dt \\&=π\int_{0}^{\pi}f(t) dt-I.\end{align}$$ Hence, we have $$2I=π\int_{0}^{\pi}f(t) dt.$$

Also, since the graph of $y=f(t)$ is symmetric with $x=π/2$, $$\int_{0}^{\pi}f(t) dt=2\int_{0}^{{\pi}/{2}} f(t) dt.$$ Hence, $$I=π\int_{0}^{{\pi}/{2}} f(t) dt.$$ Here, using $\cos t =u$, $$\begin{align}I=π\int_{1}^{0} \frac{\sin t}{{1+\cos^2t}} \left(\frac{-du}{\sin t}\right) =π\int_{0}^{1} \frac{1}{1+u^2} du=\frac{{\pi}^2}{4}. \end{align}$$ So, we know that we can get the answer you wrote.

Answer 2

There is an nice little result that says that $$\int_0^\pi xf(\sin x)dx=\frac{\pi}2\int_0^\pi f(\sin x)dx$$

Can you prove it? Then, it all comes down to using the chain rule.

Answer 3

$$ \begin{align} \int_0^\pi\frac{x\sin(x)}{1+\cos^2(x)}\,\mathrm{d}x &=-\int_0^\pi\frac{x}{1+\cos^2(x)}\,\mathrm{d}\cos(x)\tag{1}\\ &=-\int_0^\pi x\,\mathrm{d}\arctan(\cos(x))\tag{2}\\ &=-x\arctan(\cos(x))\Big]_0^\pi+\int_0^\pi\arctan(\cos(x))\,\mathrm{d}x\tag{3}\\ &=\frac{\pi^2}{4}+\int_{-\pi/2}^{\pi/2}\arctan(\sin(x))\,\mathrm{d}x\tag{4}\\ &=\frac{\pi^2}{4}\tag{5} \end{align} $$ Justification:
$(1)$: $\mathrm{d}\cos(x)=-\sin(x)\,\mathrm{d}x$
$(2)$: $\mathrm{d}\arctan(u)=\frac1{1+u^2}\mathrm{d}u$
$(3)$: integration by parts
$(4)$: substitute $x\mapsto\frac\pi2-x$
$(5)$: integral of an odd function over a symmetric domain is $0$

Answer 4

I prove lemma that says that $$\int_0^\pi xf(\sin x)dx=\frac{\pi}2\int_0^\pi f(\sin x)dx$$ let $t=\pi-x$,so \begin{align*} \int_0^\pi xf(\sin x)dx&=\int_\pi^0(\pi-t)f(\sin(\pi-t))(-dt)=\int_0^\pi(\pi-t)f(\sin t)dt\\ &=\pi\int_0^\pi f(\sin t)dt-\int_0^\pi tf(\sin t)dt \end{align*} so $$\int_0^\pi xf(\sin x)dx=\dfrac{\pi}{2}\int_0^\pi f(\sin x)dx$$ Finally \begin{align*} \int_0^{\pi} \frac{x \sin x}{1+\cos ^2 x}dx&=\dfrac{\pi}{2}\int_0^\pi\frac{ \sin x}{1+\cos ^2 x}dx=-\dfrac{\pi}{2}\int_0^{\pi}\frac{1}{1+\cos ^2 x}d\cos x\\ &=-\dfrac{\pi}{2}\arctan\cos x\Bigg|_0^\pi=\dfrac{\pi^2}{4} \end{align*}

Answer 5

HINT:

use $$I=\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$

so that $$2I=\int_a^bf(x)dx+\int_a^bf(a+b-x)dx=\int_a^b\{f(x)+f(a+b-x)\}dx$$

using $\sin(\pi+0-x)=\sin x$ and $\cos(\pi+0-x)=-\cos x$