Integral question unable to solve by primitive

Question

So I have this integral question and it seems like it's primitive it's not so appealing.... so I'm thinking there should be another way to solve it.... but I seem to can't figure out how.

$\int_0^{2\pi}x\frac{\cos x}{2-\cos^2x}dx$

I entered it on wolframalpha and it seemed like it's answer is 0. So I'm thinking this shouldn't be a coincidence and there must be a smart way to solve this.

Answer 1

Do the substitution $u = 2\pi - x$ to see that $$\int_0^{2\pi}x\frac{\cos x}{2-\cos^2x}dx = \int_0^{2\pi} (2\pi - u) \frac{\cos u}{2-\cos^2u}du = \int_0^{2\pi} (2\pi - x) \frac{\cos x}{2-\cos^2x}dx.$$ Take the average of both sides to see that $$\int_0^{2\pi}x\frac{\cos x}{2-\cos^2x}dx = \pi \int_0^{2\pi} \frac{\cos x}{2-\cos^2x}dx.$$ \begin{align} \int_0^{2\pi}x\frac{\cos x}{2-\cos^2x}dx &= \pi \int_0^{2\pi} \frac{\cos x}{2-\cos^2x}dx \\ &= \pi \int_0^{2\pi} \frac{d(\sin x)}{1+\sin^2x} \\ &= \pi \left[\arctan(\sin x)\right]_0^{2\pi} \\ &= \pi [\arctan(2\pi) - \arctan(0)] \\ &= 0 \end{align}

Answer 2

Guideline use $\int_a^b f(x)dx=\int _a^b f(a+b-x)$ then after applying it once use $1-\cos^2(x)=\sin^2(x)$ and then let $\sin(x)=u$

Answer 3

You already did integration by parts, giving you the integral $\int_0^{2\pi}\arctan(\sin x) dx$. By symmetry, this is the same as $\int_{-\pi}^{\pi}\arctan(\sin x) dx$, and that integrand is an odd function.