$\int_{-\pi}^{\pi} \frac{2x(1+\sin(x))}{1+\cos^2(x)} \mathrm d x$

Question

$$\int_{-\pi}^{\pi} \frac{2x(1+\sin(x))}{1+\cos^2(x)} \ \mathrm dx$$

My attempt:

$$\int_{-\pi}^{\pi} \frac{2x(1+\sin(x))}{1+\cos^2(x)} \ \mathrm dx =\int_{-\pi}^{\pi} \frac{2x}{1+\cos^2(x)} \ \mathrm dx+\int_{-\pi}^{\pi} \frac{2x\sin(x)}{1+\cos^2(x)} \mathrm dx$$ $$=0+\int_{-\pi}^{\pi} \frac{2x\sin(x)}{1+\cos^2(x)} \ \mathrm dx$$

I'm not sure how to proceed further. Integration by parts doesn't seem promising. Neither does any substitution.

Any suggestions?

The integral is even so you just have $2 \int_0^\pi \frac{x \sin x}{1 + \cos^2 x} \mathrm d x$. For this integral, see this question: I need assistance in integrating $ \frac{x \sin x}{1+(\cos x)^2}$. — Toby Mak, Aug 29 '20 at 06:44
Use https://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3x/439856#439856 — lab bhattacharjee, Aug 29 '20 at 06:48
@TobyMak it did help. Also how is $\frac{xsinx}{1+cos^2x}$ not even? — DatBoi, Aug 29 '20 at 06:49

score 1 · Accepted Answer · edited Aug 30 '20 at 08:50

1

Consider $$\int_{-\pi}^{\pi}\frac{2x\sin x}{1+{\cos}^2x}\,dx=2\int_{0}^{\pi}\frac{2x\sin x}{1+{\cos}^2x}\,dx$$

Now replace $x$ by $\pi-x$ and add the integrals.

edited Aug 30 '20 at 08:50

C Squared

3,679

answered Aug 29 '20 at 06:46

Hari Ramakrishnan Sudhakar

12,089

That helped. Thanks – DatBoi Aug 29 '20 at 06:49

$\int_{-\pi}^{\pi} \frac{2x(1+\sin(x))}{1+\cos^2(x)} \mathrm d x$

1 Answers1