<s>[INST] 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? 
 [/INST] Simply write 
$$\sin^2x\cos^2x=\dfrac14\sin^22x=\dfrac18-\dfrac18\cos4x$$
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