I want to solve the following IBVP:
$$u_{tt}=\alpha^2 u_{xx} \space \space \space \space \space \space 0<x<L; \space \space \space \space \space \space0<t<\infty \\ BCs: u(0,t)=0; u(L,t)=0\space \space \space \space \space \space 0<t< \infty \\ ICs:u(x,0)=\sin{\frac{\pi x}{L}}+0.5 \sin{\frac{3 \pi x}{L}}; u_t(x,0)=0$$
The solution to the finite string problem is:
$$\boxed{u(x,t)=\sum_{n=1}^{\infty}\sin{\frac{n \pi x}{L}}[a_n \sin{\frac{n \pi \alpha t}{L}}+b_n \cos{\frac{n \pi \alpha t}{L}}]}$$
where:
$$\boxed{a_n=\frac{2}{n \pi \alpha} \int_0^L g(x)\sin{\frac{n \pi x}{L}} dx}$$
$$\boxed{b_n=\frac{2}{L} \int_0^L f(x) \sin{\frac{n \pi x}{L}} dx}$$
In this case $g(x)=0$ and $f(x)=\sin{\frac{\pi x}{L}}+0.5 \sin{\frac{3 \pi x}{L}}$.
$\implies a_n=0; \space \space \space b_n=\frac{2}{L}\int_0^L(\sin{\frac{\pi x}{L}}+0.5 \sin{\frac{3 \pi x}{L}})\sin{\frac{n \pi x}{L}} dx$
I am stuck here. I don't really know how to evaluate the integral without it becoming really messy. The solution is supposed to be:
$$u(x,t)=\sin{\frac{\pi x}{L}} \cos{\frac{\pi \alpha t}{L}}+\frac{1}{2}\sin{\frac{3 \pi x}{L}}\cos{\frac{3 \pi \alpha t}{L}}$$
I don't see how I can arrive at this solution. Doesn't the solution have to be an infinite series? Am I even doing this correctly?
The $L$ is given by the boundary conditions. It can be thought of as the length of a string. Observe that $$ \begin{aligned} \cos 2\theta&=\cos^2 \theta-\sin^2 \theta\\ &=(1-\sin^2 \theta)-\sin^2 \theta\\ &=1-2\sin^2 \theta. \end{aligned} $$ Therefore $\sin^2 \theta=\frac{1}{2}-\frac{\cos 2\theta}{2}.$ When $\theta=\frac{n\pi x}{L},$ this means $$ \begin{aligned} \int_0^L(\sin \frac{n\pi x}{L})^2 dx&=\int_0^L\frac{1}{2}dx+\int_0^L -\frac{\cos \frac{2n\pi x}{L}}{2} dx\\ &=\frac{L}{2}-\frac{1}{2}\int_0^L\cos \frac{2n\pi x}{L} dx\\ &=\frac{L}{2}-\frac{1}{2}\Big[\frac{L}{2 n\pi}\sin \frac{2n\pi x}{L}\Big]_0^L\\ &=\frac{L}{2}-\frac{1}{2}(0-0)=\frac{L}{2}. \end{aligned} $$
We would like this integral to equal $1$, so we take this integral multiplied by $\frac{2}{L}$. Hence the definition of $b_n$ in your post.
Of course, for $m\ne n$, we still have $$ \int_0^L\sin \frac{n\pi x}{L}\sin \frac{m\pi x}{L} dx=0, $$ so that you only have to worry about $n=1$ and $n=3$ in your case.
The case where $L=1$ is just a (nice) example of the general case.