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.

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
$$
\frac{\sin x-\sin 3 x+\sin 5 x}{\cos x+\cos 3 x+\cos 5 x}=\tan x
$$
