Calculating values of integrals using Fourier series and uniform convergence I have a problem that I don't know how to begin solving. I have f(t)
$$
f(t) = \sum_{k=1}^\infty\frac{1}{k^2+1}\sin{kt}
$$
First I had to show that this series converges uniformly, I've done that using Weierstrass M-test comparing the series to
$$
\sum_{k=1}^\infty\frac{1}{k^2}
$$ 
Then using this result (and possibly Fourier series), I have to solve these 2 integrals
$$
\int_{-\pi}^\pi f(t)\sin{3t}dt \qquad \text and \qquad \int_{-\pi}^\pi f(t)\cos{3t}dt
$$
The problem is that I have no idea how to begin solving this. I tried just sticking $f(t)$ inside these integrals, and then using uniform convergence to swap the integral and sum, but I end up with an integral that I don't know how to solve.
Could i get some hints?
 A: You just have to use the main property (orthogonality) of the Fourier base. Provided that $a,b\in\mathbb{N}$,
$$ \int_{-\pi}^{\pi}\sin(ax)\cos(bx)\,dx=0\tag{1}$$
as well as:
$$\int_{-\pi}^{\pi}\cos(ax)\cos(bx)\,dx=\int_{-\pi}^{\pi}\sin(ax)\sin(bx)\,dx= \pi\cdot\delta(a,b)\tag{2} $$
hence your integrals equal $\frac{\pi}{10}$ and $0$ by termwise integration.
A: To continue the answer posted by @JackD'Aurizio, recall that $\sin(nx)\sin(mx)=\frac12(\cos((n-m)x)-\cos((n+m)x))$.  Then, we have
$$\int_{-\pi}^{\pi}\sin(nx)\sin(mx)\,dx=\frac12 \int_{-\pi}^{\pi}\cos((n-m)x)\,dx-\frac12 \int_{-\pi}^{\pi}\cos((n+m)x)\,dx$$
For $n=m$  
$$\int_{-\pi}^{\pi}\cos((n-m)x)\,dx=2\pi$$
while for $n\ne m$ we have
$$\begin{align}
\int_{-\pi}^{\pi}\cos((n-m)x)\,dx&=\left.\frac{\sin((n-m)x)}{n-m}\right|_{-\pi}^{\pi}\\\\
&=\frac{\sin((n-m)\pi)}{n-m}-\frac{\sin(-(n-m)\pi)}{n-m}\\\\
&=0-0\\\\
&=0
\end{align}$$
Similarly, for $n=-m$  
$$\int_{-\pi}^{\pi}\cos((n+m)x)\,dx=2\pi$$
while for $n\ne -m$ we have
$$\begin{align}
\int_{-\pi}^{\pi}\cos((n+m)x)\,dx&=\left.\frac{\sin((n+m)x)}{n+m}\right|_{-\pi}^{\pi}\\\\
&=\frac{\sin((n+m)\pi)}{n+m}-\frac{\sin(-(n+m)\pi)}{n+m}\\\\
&=0-0\\\\
&=0
\end{align}$$

Analogously, recall that $\sin(nx)\cos(mx)=\frac12(\sin((n+m)x)+\sin((n-m)x))$.  Then, 
$$\int_{-\pi}^{\pi}\sin(nx)\cos(mx)\,dx=\frac12 \int_{-\pi}^{\pi}\sin((n+m)x)\,dx-\frac12 \int_{-\pi}^{\pi}\sin((n-m)x))\,dx$$
For $n=-m$  
$$\int_{-\pi}^{\pi}\sin((n+m)x)\,dx=0$$
and for $n\ne -m$ we have
$$\begin{align}
\int_{-\pi}^{\pi}\sin((n+m)x)\,dx&=\left.\frac{-\cos((n+m)x)}{n+m}\right|_{-\pi}^{\pi}\\\\
&=\frac{-\cos((n+m)\pi)}{n+m}-\frac{-\cos(-(n+m)\pi)}{n+m}\\\\
&=-(-1)^{n+m}+(-1)^{n+m}\\\\
&=0
\end{align}$$
Similarly, for $n=m$  
$$\int_{-\pi}^{\pi}\sin((n-m)x)\,dx=0$$
and for $n\ne m$ we have
$$\begin{align}
\int_{-\pi}^{\pi}\sin((n-m)x)\,dx&=\left.\frac{-\cos((n-m)x)}{n-m}\right|_{-\pi}^{\pi}\\\\
&=\frac{-\cos((n-m)\pi)}{n-m}-\frac{-\cos(-(n-m)\pi)}{n-m}\\\\
&=-(-1)^{n-m}+(-1)^{n-m}\\\\
&=0
\end{align}$$

Putting everything together gives
$$
\int_{-\pi}^{\pi}\sin(nx)\sin(mx)\,dx=
\begin{cases}
\pi,&n=m\\\\
0,&n\ne m
\end{cases}
$$
and
$$\int_{-\pi}^{\pi}\sin(nx)\cos(mx)\,dx=0$$
