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

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.
