Evaluate $\int_{1/10}^{1/2}\left(\frac{\sin{x}-\sin{3x}+\sin{5x}}{\cos{x}+\cos{3x}+\cos{5x}}\right)^2dx$ How can I evaluate the following integral?

$$\int_{1/10}^{1/2}\left(\frac{\sin{x}-\sin{3x}+\sin{5x}}{\cos{x}+\cos{3x}+\cos{5x}}\right)^2dx$$  

I notice that the numerator and the denominator are very similar. So my direction is to evaluate this integral by substitution. However, I cannot find a suitable substitution so that the integral can be nice.
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,}
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 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}
&\color{#f00}{\sin\pars{x} - \sin\pars{3x} + \sin\pars{5x}}
\\[3mm] = &\
\Im\sum_{k = 0}^{2}\pars{-1}^{k}\expo{\ic\pars{2k + 1}x} =
\Im\bracks{\expo{\ic x}\,
{\pars{-\expo{2\ic x}}^{3} - 1 \over -\expo{2\ic x} - 1}} =
\sin\pars{3x}\,{\cos\pars{3x} \over \color{#f00}{\cos\pars{x}}}\tag{1}
\\
&\ --------------------------------
\\
&\color{#f00}{\cos\pars{x} + \cos\pars{3x} + \cos\pars{5x}}
\\[3mm] = &\
\Re\sum_{k = 0}^{2}\expo{\ic\pars{2k + 1}x} =
\Re\bracks{\expo{\ic x}\,
{\pars{\expo{2\ic x}}^{3} - 1 \over \expo{2\ic x} - 1}} =
\cos\pars{3x}\,{\sin\pars{3x} \over \color{#f00}{\sin\pars{x}}}\tag{2}
\\
&\ --------------------------------
\\ &\
\mbox{Then,}\quad {\pars{1} \over \pars{2}} = \color{#f00}{\tan\pars{x}}
\end{align}
A: Someone's suggested Chebyshev polynomials.  I hadn't thought of that, but I know the tangent half-angle formula:
$$
\frac{\sin\alpha+\sin\beta}{\cos\alpha+\cos\beta} = \tan\frac{\alpha+\beta}2.
$$
From that we get
$$
\frac{\sin(-3x) + \sin(5x)}{\cos(-3x) + \cos(5x)} = \tan x.
$$
Since $\cos(-3x)=\cos(3x)$, that's the same as the function that gets squared except it's missing the two functions of $1\cdot x$.  But if
$$
\frac a b = \frac c d
$$
then both are equal to
$$
\frac{a+b}{c+d}
$$
and that can be applied in the case where we have
$$
\frac a b = \frac{\sin(-3x) + \sin(5x)}{\cos(-3x) + \cos(5x)} = \tan x = \frac{\sin x}{\cos x} = \frac c d.
$$
with the ultimate conclusion that the fraction that gets squared under the integral sign is $\tan x$.
A: Maybe it is useful to notice that

$$ \frac{\sin x-\sin(3x)+\sin(5x)}{\cos(x)+\cos(3x)+\cos(5x)}=\color{red}{\tan x}\tag{1}$$

and $\int \tan^2 x\,dx = -x+\tan(x).$ In terms of Chebyshev polynomials, $(1)$ is equivalent to:
$$ x\cdot\left(1-U_2(x)+U_4(x)\right) = T_1(x)+T_3(x)+T_5(x)\tag{2}$$
that is straightforward to check.
A: Using product-to-sum formula, we have
$$
\begin{aligned}
2 \cos x(\sin x-\sin 3 x+\sin 5 x) = & \sin 2 x-(\sin 4 x+\sin 2 x)+(\sin 6 x+\sin 4 x) \\
= & \sin 6 x \cdots (1)
\end{aligned}
$$
and
$$
\begin{aligned}
2 \sin x(\cos x+\cos 3 x+\cos 5 x) = & \sin 2 x+(\sin 4 x-\sin 2 x)+(\sin 6 x-\sin 4 x) \\
= & \sin 6 x \cdots (2) 
\end{aligned}
$$
$(1)\div(2)$ yields
$$
\frac{\sin x-\sin 3 x+\sin 5 x}{\cos x+\cos 3 x+\cos 5 x}=\tan x
$$
