Integrate $\sin^4(4x)/\sin^2(x)$ I would like to compute
$$\int_{0}^{2\pi} \frac{\sin^4(4x)}{\sin^2(x)} \mathrm{d}x.$$
Wolfram|Alpha is able to compute an antiderivative explicitly so I do not think use of the residue theorem is needed, but I'm interested in any approach.
 A: We can compute the integral by Fourier series. Consider
\begin{align}
f(x) = \frac{\sin^2(4x)}{\sin x}
\end{align}
then the sine series expansion of $f$ is given by
\begin{align}
f(x) = \sin x + \sin 3x + \sin 5x + \sin 7 x.
\end{align}
In particular, we see
\begin{align}
\int^{2\pi}_0 f^2(x)\ dx = \int^{2\pi}_0 \sin^2 x+ \sin^2(3x) + \sin^2(5x) + \sin^2(7x)\ dx 
\end{align}
since
\begin{align}
\int^{2\pi}_0 \sin(nx) \sin(mx) \ dx = 0 \ \ \text{ if } \ \ n\neq m. 
\end{align}
Lastly, we have that
\begin{align}
\int^{2\pi}_0 \sin^2(n x)\ dx = \frac{1}{2}\int^{2\pi}_0 1-\cos(2nx)\ dx = \pi. 
\end{align}
Thus, it follows
\begin{align}
\int^{2\pi}_0 \frac{\sin^4(4x)}{\sin^2 x}\ dx = 4\pi. 
\end{align}
Remark: The hard work lies in finding the sine series expansion. 
A: One approach: use the double-angle formulas
$$\sin(2a) = 2 \sin(a) \cos(a)$$
and
$$\cos(2a) = 1 - 2 \sin^2(a).$$
Specifically, rewrite the numerator as
$$\sin^4(4x) = [2 \sin(2x) \cos(2x)]^4 = [4 \sin(x) \cos(x) (1 - 2 \sin^2(x))]^4.$$
It's not particularly pretty from there, but at least your denominator term will get fully cancelled....
A: $$\frac{\sin(4x)}{\sin(x)}=\frac{e^{4ix}-e^{-4ix}}{e^{ix}-e^{-ix}}=e^{3ix}+e^{ix}+e^{-ix}+e^{-3ix}=2\cos(x)+2\cos(3x) $$
$$\frac{\sin^2(4x)}{\sin(x)}=\sin(x)+\sin(3x)+\sin(5x)+\sin(7x) $$
hence by Parseval's theorem
$$ \int_{0}^{2\pi}\frac{\sin^4(4x)}{\sin^2(x)}\,dx = \color{red}{4\pi}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\int_{0}^{2\pi}{\sin^{4}\pars{4x} \over \sin^{2}\pars{x}}\,\dd x & =
\oint_{\verts{z}\ =\ 1^{+}}{\bracks{\pars{z^{4} - 1/z^{4}}/\pars{2\ic}}^{\, 4} \over
\bracks{\pars{z - 1/z}/\pars{2\ic}}^{\, 2}}\,{\dd z \over \ic z} =
-\,{1 \over 4\ic}
\oint_{\verts{z}\ =\ 1^{+}}{\pars{1 - z^{8}}^{4} \over
\pars{1 - z^{2}}^{2}}\,{\dd z \over z^{15}}
\\[5mm] & =
-\,{1 \over 4\ic}
\sum_{k = 0}^{4}\sum_{n = 0}^{\infty}{4 \choose k}\pars{-1}^{k}{-2 \choose n}
\pars{-1}^{n}\oint_{\verts{z}\ =\ 1^{+}}{\dd z \over z^{15 - 8k - 2n}}
\\[5mm] & =
-\,{1 \over 4\ic}
\sum_{k = 0}^{4}\sum_{n = 0}^{\infty}{4 \choose k}\pars{-1}^{k}{n + 1 \choose n}
\braces{\vphantom{\Large A}2\pi\ic\bracks{15 - 8k - 2n = 1}}
\\[5mm] & =
-\,{\pi \over 2}
\sum_{k = 0}^{4}\sum_{n = 0}^{\infty}{4 \choose k}\pars{-1}^{k}
\pars{n + 1}\bracks{n = 7 - 4k}
\\[5mm] & =
-\,{\pi \over 2}\sum_{k = 0}^{4}{4 \choose k}\pars{-1}^{k}
\bracks{\pars{7 - 4k} + 1}\bracks{7 - 4k \geq 0}
\\[5mm] & =
-2\pi\sum_{k = 0}^{4}{4 \choose k}\pars{-1}^{k}
\pars{2 - k}\bracks{k \leq {7 \over 4}} =
-2\pi\sum_{k = 0}^{1}{4 \choose k}\pars{-1}^{k}\pars{2 - k}
\\[5mm] & = \bbx{4\pi}
\end{align}
