Integral of $\int \frac{\sin(3x)}{\sin(5x)} \, dx$ How to find integral of

$$\int \frac{\sin(3x)}{\sin(5x)}dx$$

I wrote $\sin(3x)=\sin(8x-5x)$ but it generated $\frac{\sin(8x) \cos(5x)}{\sin(5x)}$. 
How should I proceed?
 A: Use $\;\sin3x=\sin(5x-2x)=\sin5x\cos2x-\sin2x\cos5x$ :
$$\frac{\sin3x}{\sin5x}=\cos 2x-\sin2x\frac{\cos5x}{\sin5x}$$
Now, observe that
$$\int\frac{\cos kx}{\sin kx}dx=\frac1k\,\log|\sin kx|+C$$
and now perhaps integrating by parts will help.
A: Using Chebyshev polynomials (that can be used also for sine when you have odd multipliers for the angle) we have: $$
\sin3x = 3\sin x-4\sin^3x \\ 
\sin5x = 16\sin^5x-20\sin^3x+5\sin x
$$
So: $$
\frac{\sin3x}{\sin5x}=\frac{3-4\sin^2x}{16\sin^4x-20\sin^2x+5}=\frac{-\left (4\sin^2x-\frac{5}{2} \right )+\frac{1}{2}}{\left (4\sin^2x-\frac{5}{2} \right )^2-\frac{5}{4}}
$$ Let's convert sines to cosines (they look better when you integrate): $$
\frac{\sin3x}{\sin5x}=\frac{-\left (4(1-\cos^2x)-\frac{5}{2} \right )+\frac{1}{2}}{\left (4(1-\cos^2x)-\frac{5}{2} \right )^2-\frac{5}{4}}=\frac{\left (4\cos^2x-\frac{3}{2} \right )+\frac{1}{2}}{\left (4\cos^2x-\frac{3}{2} \right )^2-\frac{5}{4}}
$$ Now we can decompose the denominator: $$
\frac{u+\frac{1}{2}}{u^2-\frac{5}{4}}=\frac{a}{u-\sqrt{\frac{5}{4}}}+\frac{b}{u+\sqrt{\frac{5}{4}}} \\ a=\frac{\sqrt{5}+1}{2\sqrt{5}}, b=\frac{\sqrt{5}-1}{2\sqrt{5}}
$$ Then: $$
\frac{\sin3x}{\sin5x}=\frac{\frac{\sqrt{5}+1}{2\sqrt{5}}}{4\cos^2x-\frac{3}{2}-\sqrt{\frac{5}{4}}}+\frac{\frac{\sqrt{5}-1}{2\sqrt{5}}}{4\cos^2x-\frac{3}{2}+\sqrt{\frac{5}{4}}}
$$ For both the denominators we have a negative constant term, so let's try to find a solution for: $$
I(a^2, b^2)=\int {\frac{1}{a^2cos^2x-b^2}}dx=\int {\frac{1+\tan^2x}{a^2-b^2-b^2\tan^2x}}dx$$ using $u=\tan x$, $du=(1+\tan^2x)dx$: $$
I(a^2, b^2)=\int {\frac{1}{a^2-b^2-b^2u^2}}du=\frac{1}{b^2}\int {\frac{1}{\frac{a^2-b^2}{b^2}-u^2}}du=\frac{1}{b^2}\frac{b}{\sqrt{a^2-b^2}}\tanh^{-1}\frac{bu}{\sqrt{a^2-b^2}}=\frac{1}{b\sqrt{a^2-b^2}}\tanh^{-1}\frac{b\tan x}{\sqrt{a^2-b^2}}
$$ Carefully doing the substitutions I found: $$
\int {\frac{\sin3x}{\sin5x}}dx=   I\left (4, \frac{3+\sqrt{5}}{2}\right )   + I\left (4, \frac{3-\sqrt{5}}{2}\right )=   \frac{\sqrt{5}+1}{4\sqrt{5}}\frac{1}{\sqrt{3+\sqrt{5}}\sqrt{5-\sqrt{5}}}\tanh^{-1}\frac{\sqrt{3+\sqrt{5}}}{\sqrt{5-\sqrt{5}}}\tan x+\frac{\sqrt{5}-1}{4\sqrt{5}}\frac{1}{\sqrt{3-\sqrt{5}}\sqrt{5+\sqrt{5}}}\tanh^{-1}\frac{\sqrt{3-\sqrt{5}}}{\sqrt{5+\sqrt{5}}}\tan x + C
$$
A: Express
\begin{align}
\sin(3x)& =2\sin x  \left(\cos(2x)- \cos\frac{2\pi}{3}\right) \\
\sin(5x)& =4\sin x  \left(\cos(2x)- \cos\frac{2\pi}{5}\right) \left(\cos(2x)- \cos \frac{4\pi}{5}\right)
\end{align}
Then, decompose the integrand as follows
\begin{align}\frac{\sin (3x)}{\sin(5x)}
=&\frac{\cos(2x)- \cos\frac{2\pi}{3}} {2\left(\cos(2x)- \cos\frac{2\pi}{5}\right) \left(\cos(2x)- \cos \frac{4\pi}{5}\right)}\\
=&\frac{\frac1{\sqrt5}\cos\frac{2\pi}{5} }{\cos(2x)- \cos\frac{4\pi}{5} }
-\frac{ \frac1{\sqrt5} \cos\frac{4\pi}{5}}{\cos(2x)- \cos\frac{2\pi}{5} }
\\
=&\frac{\frac1{2\sqrt5}\cos\frac{2\pi}{5} }{\sin(\frac{2\pi}5+x)\sin(\frac{2\pi}5-x) }-\frac{\frac1{2\sqrt5} \cos\frac{4\pi}{5} }{\sin(\frac{\pi}5+x)\sin(\frac{\pi}5-x) }\\
=&\frac15\sin\frac\pi5\left(\cot(\frac{2\pi}5+x)+\cot(\frac{2\pi}5-x) \right)\\
&\hspace{1cm}+\frac15\sin\frac{2\pi}5\left(\cot(\frac{\pi}5+x)+\cot(\frac{\pi}5-x) \right)
\end{align}
Apply $(\ln \sin t)’= \cot t$ to integrate
$$\int \frac{\sin (3x)}{\sin(5x)}dx= \frac15\sin\frac{\pi}5\>\ln \frac{\sin(\frac{2\pi}5+x) }{\sin(\frac{2\pi}5-x) } +\frac15\sin\frac{2\pi}5\>\ln \frac{\sin(\frac{\pi}5+x) }{\sin(\frac\pi5-x) }+C
$$
