Find the limit of the sequence $(b_n)^\infty _{n=1}$ where $b_n$ is the integral from $-2\pi n$ to $\pi + 2\pi n$ of $x\cos(x)$ I am struggling to complete this question. I know that the function is symmetrical, so the integral from $-n$ to $n$ of $x\cos(x)$ is just 0. but this has an additional $\pi$. I tried splitting up the integral to two, one going from $-2\pi n$ to $2\pi n$ and another from $2\pi n$ to $2\pi n + \pi$ and that just doesn't work, its just as hard to solve. Is the only way to just solve normally? It just will be a very very long process so I am suspicious.

Note that $\int _{a}^bx\cos(x)dx=
a\sin(a)-\cos(a)-(b\sin(b)-\cos(b)$) which can be derived by integration by parts and then using the fundamental theorem of calculus .
