Suppose that we have the problem $\begin{cases}u_{tt} - c^2u_{xx} = 0&: x \in (0, \pi), t > 0\\\ u_t(x, 0) = \sin(10x), &: x \in [0, \pi]\\\ u(0, t) = u(\pi, t) = 0, &: t \in [0, \infty)\\\ u(x, 0) = \sin^5(x),&: x \in [0, \pi]\end{cases}$
I know that the general solution of the wave equation in my case is given by $u(x, t) = \sum_{l\in \mathbb{N}_0}\sin(lx)\left(a_l\cos(clt) + b_l\sin(clt)\right)$ where $a_l = \frac{2}{\pi}\int_{0}^\pi \sin^5(x)\sin(lx)dx$ and $b_l = \frac{2}{cl\pi}\int_{0}^\pi \sin(10x)\sin(lx)dx$
What I cannot comprehend is that if I slap the equation of $a_l$ into Wolfram|Alpha, I get the nerve wrecking result in the image below 
That is, for all $l \in \mathbb{N}$, $a_l = \frac{-240\sin(\pi l)}{\pi(l^6 - 35l^4 + 259l^2 - 225)} = 0$ as $\sin(l\pi) = 0, \forall l \in \mathbb{Z}$. But then, doesn't this imply that $u(x, 0) = \sin^5(x) = \sum_{l\in \mathbb{N}_0}a_l\sin(lx) = 0$ so that $\sin \equiv 0$?
There is clearly something wrong with my reasoning, but I have no idea where the fault is.
Good question! Notice that the polynomial $$l^6-35l^4+259l^2-225$$ Has roots at $l=\pm 1,\pm 3,\pm 5$. And, since it is a sixth order polynomial, these are the only roots. That means when you plug these values into the formula for $a_l$, you get a $0/0$. So the formula breaks down in this case. What we need to do is rewrite it as $$a_n=\frac{240}{\pi}\frac{\sin(\pi n)}{(n-1)(n+1)(n-3)(n+3)(n-5)(n+5)}$$ And due to the limit (easy exercise!) $$\lim_{x\to n}\frac{\sin(\pi x)}{n-x}=(-1)^{n+1}\pi$$ You can get for example $$a_1=\frac{240}{\pi}\frac{1}{(1+1)(1-3)(1+3)(1-5)(1+5)}\lim_{x\to 1}\frac{\sin(\pi x)}{1-x} \\ =\frac{240}{2\cdot(-2)\cdot4\cdot(-4)\cdot6}=\frac{5}{8}$$ The same process can be repeated for the other five relevant integers. And, as well we clearly get $$a_n=0 \\ \text{if}~n\neq \pm 1,\pm 3,\pm 5$$