Help with a trignometric double integral I need help understanding the evaluation of the following double integral:
$$\int_0^1 \int_0^1 x\cos(\pi y)\sin(n\pi x)\cos(m\pi y)\,dy\,dx =
     \begin{cases}
       0 & \text{if} \quad m \ne 1\\[10pt]
       \dfrac{(-1)^{n+1}}{2n\pi} & \text{if} \quad m = 1
     \end{cases}
$$
I expect that the answer lies in the property that the sine and cosine functions integrate nicely on periods of $\pi$.

First observe that $$ \int_0^1 \cos(m\pi y)\cos(n \pi y)=\begin{cases}\frac{1}{2}, & \text{if } m\neq n\\ 0, &\text{otherwise} \end{cases} .$$
\begin{align*}
\int_0^1\int_0^1 x\cos(\pi y)\sin(n\pi x)\cos(m\pi y)dydx &=\int_0^1x\ \sin(n\pi x)dx\int_0^1\cos(\pi y)\cos(m\pi y)dy\\
\end{align*}
$$ =\begin{cases}  \int_0^1x\sin (n\pi x)dx, & m=1\\ 0, & m\neq 1. \end{cases} $$ Now evaluate $ \int_0^1x\sin(n\pi x) dx $ by the help of integration by parts.
