$$ \int_{0}^{2\pi}\lvert\sin(x)\rvert\cos(nx)\,dx= -\frac{4\cos^2\bigl(\frac{\pi n}{2}\bigr)\cos(\pi n)}{n^2-1} $$
I'm not actually trying to solve this myself. The answer appears in my lecture notes without any explanation whatsoever. Apparently it depends on whether or not $n$ is odd or even, since the answer is $0$ when $n$ is odd, but I really don't understand how they've gotten the answer at all. Please help me to figure out what the heck is going on!
On
the graph of $y = |\sin(x)|\cos(nx)$ is $2\pi$-periodic, even and odd about $x = \pi/2$ when $n$ is odd. therefore $$\int_0^{2\pi} |\sin(x)|\cos(nx) = 0 \text{ for } n \text{ odd. }$$
in the case $n$ even, $y = |\sin(x)|\cos(nx)$ is $\pi$-periodic,and even about $x = \pi/2.$ therefore
$$\begin{align}\int_0^{2\pi} |\sin(x)|\cos(nx) &= 4 \int_0^{\pi/2} sin x \cos nx \, dx=2\int_0^{\pi/2} \left(\sin(nx + x)-\sin(nx-x)\right)\, dx\\
&= 2\left(\frac 1{n+1}\cos(n+1)x-\frac 1{n-1}\cos(n-1)x\right)\Big|_0^{\pi/2}\\
&=2\left(\frac 1{n+1}\cos(n+1)\pi/2-\frac 1{n-1}\cos(n-1)\pi/2\right) - 2\left(\frac 1{n+1}-\frac 1{n-1}\right)\\&=\frac 4{n^2 -1}\end{align}$$
It might help to divide the integral into two, $\int_0^{\pi}$ and $\int_{\pi}^{2\pi}$. In the first one, $|\sin x|=\sin x$ and in the second one $|\sin x|=-\sin x$. Then one integrate by parts two times, or better, uses the formula $$ \sin x\cos nx=\frac{1}{2}(\sin(n+1)x-\sin(n-1)x). $$ It is a good exercise to do the calculations to see how it works (and what you get).