I'm a bit of a sucker for brute force calculations. Say I want to calculate a coefficient with Fourier theory, in my case \begin{align*} a_n = \int_0^1 \sin (3\pi x) \cos (n\pi x) dx. \end{align*} Clearly, only $a_3$ is nonzero. But if I calculate this integral explicitly, I get \begin{align*} a_n = \frac{3(-1)^{n+1} - 3}{\pi(n^2-9)}. \end{align*} I'm not immediately seeing that only $n=3$ will yield a nonzero result here... So how can I evaluate the integral so that only $a_3$ remains?
EDIT: Some extra notes on my calculation. We have \begin{align*} I = \int \sin(3\pi x)\cos(n\pi x) dx = \int u \frac{dv}{dx} dx, \end{align*} where \begin{align*} u = \sin(3\pi x) \Rightarrow \frac{du}{dx} = 3\pi \cos(3\pi x),\\ \frac{dv}{dx} = \cos(n\pi x) \Rightarrow v = \frac1{n\pi}\sin(n\pi x), \end{align*} such that \begin{align*} I &= \sin(3\pi x) \cdot \frac1{n\pi}\sin(n\pi x) - \int 3\pi \cos(3\pi x) \cdot \frac1{n\pi}\sin(n\pi x) dx \\ &= \frac1{n\pi}\sin(3\pi x)\sin(n\pi x) - \frac 3n\int \cos(3\pi x)\sin(n\pi x) dx \\ &= \frac1{n\pi}\sin(3\pi x)\sin(n\pi x) - \frac 3n J. \end{align*} Now repeat the calculation with $J$, \begin{align*} J = \int \cos(3\pi x)\sin(n\pi x) dx \end{align*} where \begin{align*} u = \cos(3\pi x) \Rightarrow \frac{du}{dx} = -3\pi \sin(3\pi x),\\ \frac{dv}{dx} = \sin(n\pi x) \Rightarrow v = -\frac1{n\pi}\cos(n\pi x), \end{align*} such that \begin{align*} J &= \cos(3\pi x) \cdot \left(-\frac1{n\pi}\cos(n\pi x)\right) - \int 3\pi \sin(3\pi x) \cdot \frac1{n\pi}\cos(n\pi x) dx \\ & = -\frac1{n\pi}\cos(3\pi x) \cos(n\pi x) - \frac3n \int \sin(3\pi x) \cos(n\pi x) dx \\ & = -\frac1{n\pi}\cos(3\pi x) \cos(n\pi x) - \frac3n I \end{align*} Plugging in $J$ in the expression for $I$, we get \begin{align*} I &= \frac1{n\pi}\sin(3\pi x)\sin(n\pi x) - \frac 3n J \\ &= \frac1{n\pi}\sin(3\pi x)\sin(n\pi x) - \frac 3n \left[-\frac1{n\pi}\cos(3\pi x) \cos(n\pi x) - \frac3n I\right] \\ &= \frac1{n\pi}\sin(3\pi x)\sin(n\pi x) + \frac3{n^2\pi}\cos(3\pi x) \cos(n\pi x) + \frac9{n^2} I \\ I &= \frac{\frac1{n\pi}\sin(3\pi x)\sin(n\pi x) + \frac3{n^2\pi}\cos(3\pi x) \cos(n\pi x)}{1-\frac9{n^2}} \\ &= \frac1\pi \frac{n\sin(3\pi x)\sin(n\pi x) + 3\cos(3\pi x) \cos(n\pi x)}{n^2-9}. \end{align*} On the interval $0 < x < 1$, we then get \begin{align*} I \Big|_{0}^1 &= \frac1\pi \frac{n\sin(3\pi)\sin(n\pi) + 3\cos(3\pi) \cos(n\pi)}{n^2-9} - \frac1\pi \frac{n\sin(0)\sin(0) + 3\cos(0) \cos(0)}{n^2-9}\\ &= \frac1\pi \frac{3\cos(3\pi) \cos(n\pi)}{n^2-9} - \frac1\pi \frac{3}{n^2-9} \\ &= \frac1\pi \frac{-3(-1)^n - 3}{n^2-9}, \end{align*} where in the last step I used $\cos(n\pi) = (-1)^n$.
Considering $$a_n = \int_0^1 \sin (3\pi x) \cos (n\pi x)\, dx$$ we can make the calculations faster using $$\sin(a) \cos(b)=\frac 12\big(\sin(a+b)+\sin(a-b)\big)$$ So $$2a_n=\int_0^1 \sin\big((3+n)\pi x\big)\, dx+\int_0^1 \sin\big((3-n)\pi x\big)\, dx$$ which already shows that, for $n=3$, the second integral disappears.
So $$a_3=\frac{1+\cos (3\pi )}{2 \pi (n+3)}=0$$ For the other cases $(n\neq 3)$, after simplifications, $$a_n=-\frac{3 (1+\cos (\pi n))}{\pi \left(n^2-9\right)}$$ which implies $$a_{2n+1}=0 \quad \quad \quad a_{2n}=-\frac{6}{\pi \left(4 n^2-9\right)}$$