5
$\begingroup$

I am stuck on the integral.

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

The hint was to swap limits and add something to cancel the $x$ out? If the denominator was $1 - \cos^2x$ we'd be in business lol.

$\endgroup$
15
  • 2
    $\begingroup$ This integral has a horrible exact form, and evaluating it (according to Wolfram) from $0$ to $\pi$ gives $\pi^2/4$ $\endgroup$
    – FShrike
    Aug 18, 2021 at 19:06
  • 1
    $\begingroup$ If there was no $\;x\;$ in the numerator, as they hint, you'd be in business as you'd have an integral of the form $\;\int\frac{f'}{1+f^2} dx=\arctan f\;$ ...in fact, having$\;1-\cos^2x=\sin^2x\;$ in the denominator would render a rather horrible integral... $\endgroup$
    – DonAntonio
    Aug 18, 2021 at 19:07
  • $\begingroup$ @FShrike but how do do it without Wolfram lol I'm studying for GRE subject math this came up on U of Chi practice probems $\endgroup$
    – homosapien
    Aug 18, 2021 at 19:09
  • 1
    $\begingroup$ Keep the following in mind: GRE problems are by and large not overly difficult. They may however involve a trick to reduce them to something much simpler. Symmetries (such as $x\mapsto \pi-x$) are a good thing to try if you think the problem is very difficult at face value. $\endgroup$ Aug 18, 2021 at 19:53
  • 1
    $\begingroup$ Does this answer your question? Evaluate $\int^{\pi}_0\frac{x\sin(x)}{1+\cos^2(x)}dx$ $\endgroup$
    – Axion004
    Aug 18, 2021 at 20:15

1 Answer 1

14
$\begingroup$

Denote by $$I := \int_0^\pi \frac{x \sin(x)}{1+\cos(x)^2}dx. $$

With the change of variables $x = \pi - t$, we get that $$I = \int_0^\pi \frac{(\pi - t) \sin(\pi - t)}{1 + \cos(\pi-t)^2}dt. $$

As $\sin(\pi-t) = \sin(t)$ and as $\cos(\pi - t) = -\cos(t)$, we have that $$I = \int_0^\pi \frac{(\pi - t) \sin(t)}{1 + \cos(t)^2}dt = \pi \int_0^\pi \frac{\sin(t)}{1 + \cos(t)^2}dt - I, $$ and so $$I = \frac{\pi}{2} \int_0^\pi \frac{\sin(t)}{1 + \cos(t)^2}dt. $$

The last integral can be computed using the change of variable $\cos(t) = y$, as mentioned in the comments.

$\endgroup$
12
  • 2
    $\begingroup$ $dx=-dt$ so I think the sign of the entire answer must be flipped $\endgroup$
    – FShrike
    Aug 18, 2021 at 19:25
  • 7
    $\begingroup$ It is true that $dx = -dt$, but the change of variables $x = \pi - t$ flips the limits of integration, and these two cancel out. $\endgroup$
    – C_M
    Aug 18, 2021 at 19:26
  • 1
    $\begingroup$ Aha! Then this is a good answer with good technique. I wonder why it can't be generalised and Wolfram's monster is necessary for a general integral $\endgroup$
    – FShrike
    Aug 18, 2021 at 19:29
  • 2
    $\begingroup$ It can be generalized to $$\int_a^b f(t)dt = \int_a^b f(a+b-t)dt. $$ $\endgroup$
    – C_M
    Aug 18, 2021 at 19:29
  • 2
    $\begingroup$ I will be sure to use this technique in the future. Thank you $\endgroup$
    – FShrike
    Aug 18, 2021 at 19:31

Not the answer you're looking for? Browse other questions tagged .