# GRE subject math tough integral [closed]

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.

• This integral has a horrible exact form, and evaluating it (according to Wolfram) from $0$ to $\pi$ gives $\pi^2/4$ Aug 18, 2021 at 19:06
• 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... Aug 18, 2021 at 19:07
• @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 Aug 18, 2021 at 19:09
• 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. Aug 18, 2021 at 19:53
• Does this answer your question? Evaluate $\int^{\pi}_0\frac{x\sin(x)}{1+\cos^2(x)}dx$ Aug 18, 2021 at 20:15

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.

• $dx=-dt$ so I think the sign of the entire answer must be flipped Aug 18, 2021 at 19:25
• It is true that $dx = -dt$, but the change of variables $x = \pi - t$ flips the limits of integration, and these two cancel out.
– C_M
Aug 18, 2021 at 19:26
• 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 Aug 18, 2021 at 19:29
• It can be generalized to $$\int_a^b f(t)dt = \int_a^b f(a+b-t)dt.$$
– C_M
Aug 18, 2021 at 19:29
• I will be sure to use this technique in the future. Thank you Aug 18, 2021 at 19:31