Integrate a polynomial times sine twice by parts

For $n\ne0$,

$\begin{aligned}&I\\ & =\left[-\frac{x^2\cos nx}{n}\right]_0^\pi\\ &{}+\frac2n\int_0^\pi x\cos nx\,dx.\end{aligned}$

The remaining integral is $[x\sin(nx)/n]_0^\pi-\tfrac1n\int_0^\pi\sin nx\,dx$. Therefore $I=-\pi^2\cos(n\pi)/n+2\pi\sin(n\pi)/n^2+2[\cos(n\pi)-1]/n^3$.

For a nonzero even integer, $\cos(n\pi)=1$ and $\sin(n\pi)=0$, giving $I=-\pi^2/n$, so $k=-1$. The printed statement does not exclude the even integer $n=0$: its integral is $0$, but the requested quotient is undefined there. The displayed quotient answer assumes $n\ne0$.