Fourier Series Shortcut for $\sin(3x)\cos^{2}(7x)$

Question

Let $f: \mathbb R \to \mathbb R$, $f(x):=\sin(3x)\cos^{2}(7x)$. I am meant to find the fourier series.

We know that $(Ff)(x)=\frac{a_{0}}{2}+\sum^{\infty}_{k=1}(a_{k}\cos(kx)+b_{k}\sin(kx)), \forall x \in \mathbb R$

where $a_{k}=\frac{1}{\pi}\int^{2\pi}_{0}f(x)\cos(kx)dx$ and $b_{k}=\frac{1}{\pi}\int^{2\pi}_{0}f(x)\sin(kx)dx$

How can I integrate $\int^{2\pi}_{0}\sin(3x)\cos^{2}(7x)\cos(kx)dx$, for example? Or is there a shortcut I'm not seeing. Any hints would be helpful.

Answer 1

First, notice that your function is odd, so the expansion should only include $\sin$ terms, immediately halving the calculations you have to do.

Now, for computing $$ \frac1\pi\int_0^{2\pi}\sin(3x)\cos^2(7x)\sin(kx)\mathrm dx $$ you can rewrite $$ \cos^2(7x)=\frac12(1+\cos(14x)) $$ so really we must compute $$ \frac{1}{2 \pi}\int_0^{2\pi}\sin(3x)\sin(kx)\mathrm dx+\frac{1}{2\pi}\int_0^{2\pi}\sin(3x)\sin(kx)\cos(14x)\mathrm dx $$ on the first, use orthogonality of trig functions. On the second, use that $$ \sin(3x)\sin(kx)=\frac12(\cos(3x-kx)-\cos(3x+kx)) $$ and once again, orthogonality of trig functions is your friend.

Answer 2

HINT

Recall that

$$\cos^2\frac{\theta}{2}=\frac{1+\cos\theta}{2}\implies \cos^2 (7x)=\frac{1+\cos(14x)}{2}$$

then

$$\sin(3x)\cos^{2}(7x)=\frac12\sin (3x)+\frac12\sin(3x)\cos(14x)$$

and then use

$$\sin \theta \cos \varphi = \frac12{{\sin(\theta + \varphi) + \frac12\sin(\theta - \varphi)} }$$