Fourier series of $f(x)=\sin^2x\cos^2x$ at $(-\pi,\pi)$

Question

Find Fourier series of $f(x)=\sin^2 x \cos^2 x$ at $(-\pi,\pi)$

$f(x)$ is even so we only have to evaluate $a_0,a_n$

$$ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} \sin^2 x \cos^2 x dx = \frac{1}{4\pi} \int_{-\pi}^{\pi}\sin^2(2x) = \frac{1}{4\pi} \int_{-\pi}^{\pi}\frac{1-\cos 2x}{2} dx = \frac{1}{4} $$ and $$ \begin{split} a_n &= \frac{1}{\pi} \int_{-\pi}^{\pi}\sin^2 x \cos^2 x \cos(nx)dx\\ &= \frac{1}{4\pi} \int_{-\pi}^{\pi}\frac{\cos(nx)-\cos(2x)\cos(nx)}{2}dx\\ &=\frac{-1}{16\pi}\int_{-\pi}^{\pi}(\cos((n+2)x)+\cos((n-2)x))dx\\ &=0 \end{split} $$

So $f(x)\approx\frac{1}{8}$

Is it correct?

Answer 1

As a more effective way we have that

$$ \begin{split} f(x) &= \sin^2 x \cos^2 x=\frac12(1-\cos 2x)\cdot \frac12(1+\cos 2x) \\ &=\frac14(1-\cos^2 2x)=\frac14\left[1-\frac12\left(1+\cos 4x\right)\right]=\frac18-\frac18\cos 4x \end{split} $$

Using the same expression in the integral obviously we obtain

$$a_n= \frac{1}{\pi} \int_{-\pi}^{\pi}\sin^2 x \cos^2 x \cos(nx)dx= \frac{1}{\pi} \int_{-\pi}^{\pi} \cos(nx)\left(\frac18-\frac18\cos 4x\right)dx=$$

$$= \frac{1}{8\pi}\int_{-\pi}^{\pi} \cos(nx)dx-\frac{1}{8\pi}\int_{-\pi}^{\pi} \cos(nx)\cos 4xdx$$

which is equal to $-\frac18$ for $n=4$ and equal to $0$ otherwise.

Answer 2

Simply write $$\sin^2x\cos^2x=\dfrac14\sin^22x=\dfrac18-\dfrac18\cos4x$$