First of all, we use De Moivres’ Theorem to express $\sin 5x $ in terms of $\cos x$ and $\sin x$. $$ \begin{aligned} &\cos 5 x+i \sin 5 x\\=&(\cos x+i \sin x)^{5} \\ =& \cos^{5} x+5 i \cos ^{4} x \sin x-10 \cos^{3} x \sin ^{2} x-10 i \cos ^{2} x \sin ^{3} x+5 \cos x \sin ^{4} x+i \sin ^{5} x \end{aligned} $$ Comparing the imaginary parts on both sides yields $$ \sin 5 x=5 \cos 4 \sin x-10 \cos ^{2} x \sin ^{3} x+\sin ^{5} x $$
$$ \begin{aligned} \because \int\frac{\sin x}{\sin 5 x} d x &=\int \frac{1}{5 \cos ^{4} x-10 \cos ^{2} x \sin ^{2} x+\sin ^{4} x} d x \\ &=\int \frac{\sec ^{4} x d x}{5-10 \tan ^{2} x+\tan ^{4} x} \\ &=\int \frac{1+t^{2}}{t^{4}-10 t^{2}+5}dt, \quad \textrm{ where }t=\tan x. \end{aligned} $$
Playing a small trick on the integrand yields
$$ \begin{aligned} \int \frac{\sin x}{\sin 5 x}dx&=\int \frac{1+\frac{1}{t^{2}}}{t^{2}+\frac{5}{t^{2}}-10} d t\\ &= \int \frac{\frac{\sqrt{5}+1}{2}\left(1+\frac{\sqrt{5}}{t^{2}}\right)+\frac{\sqrt{5}-1}{2}\left(1-\frac{\sqrt{5}}{t^{2}}\right)}{t^{2}+\frac{5}{t}-10} d t\\ &=\frac{\sqrt{5}+1}{2} \int \frac{d\left(t-\frac{\sqrt{5}}{t}\right)}{\left(t-\frac{\sqrt{5}}{t}\right)^{2}-(10-2 \sqrt{5})}+\frac{\sqrt{5}-1}{2} \int \frac{d\left(t+\frac{\sqrt{5}}{t}\right)}{\left(t+\frac{\sqrt{5}}{t}\right)^{2}-(10+2 \sqrt{5})}\\&=\frac{\sqrt{5}+1}{2 \sqrt{10-2 \sqrt{5}}} \tan ^{-1}\left(\frac{t-\frac{\sqrt{5}}{t}}{\sqrt{10-2 \sqrt{5}}}\right)+\frac{\sqrt{5}-1}{4 \sqrt{10+2 \sqrt{5}}} \ln \left|\frac{t+\frac{\sqrt{5}}{t}-\sqrt{10+2 \sqrt{5}}}{t+\frac{\sqrt{5}}{t}+\sqrt{10+2 \sqrt{5}}}\right|+C\\&=\frac{\sqrt{5}+1}{2 \sqrt{10-2 \sqrt{5}}} \tan ^{-1}\left(\frac{\tan^2 {x}-\sqrt{5} }{\tan {x}\sqrt{10-2 \sqrt{5}}}\right)\\&+\frac{\sqrt{5}-1}{4 \sqrt{10+2 \sqrt{5}}} \ln \left| \frac{\tan ^{2} x-\sqrt{10+2 \sqrt{5}} \tan x+ \sqrt{5}}{\tan ^{2} x+\sqrt{10+2 \sqrt{5}} \tan x+\sqrt{5}}\right|+C \end{aligned} $$
Your comments and alternative solutions are highly appreciated.
In terms of the Gaussian hypergeometric function : if $$I_n=\int \frac{\sin(x)}{\sin(nx)}\,dx$$then $$I_n=\frac{i e^{i (n+1) x} \, _2F_1\left(1,\frac{n+1}{2 n};\frac{1}{2} \left(3+\frac{1}{n}\right);e^{2 i n x}\right)}{n+1}-\frac{i e^{i (n-1) x} \, _2F_1\left(1,\frac{n-1}{2 n};\frac{1}{2} \left(3-\frac{1}{n}\right);e^{2 i n x}\right)}{n-1}$$