Wolfram Mathematica $9.0$ is able to evaluate this indefinite integral. Here is the output
\begin{equation}
\sqrt{\frac{5+\sqrt{5}}{2}}{\rm{artanh}}\left(\frac{\left(\sqrt{5}-3\right)\tan x}{\sqrt{10-2\sqrt{5}}}\right)+\sqrt{\frac{5-\sqrt{5}}{2}}{\rm{artanh}}\left(\frac{\left(\sqrt{5}+3\right)\tan x}{\sqrt{10+2\sqrt{5}}}\right)+C
\end{equation}
The complete proof can be downloaded here.
Using the facts
\begin{align}
\sin\left(\frac{\pi}{5}\right)&=\frac{1}{2}\sqrt{\frac{5-\sqrt{5}}{2}}\\[10pt]
\cos\left(\frac{\pi}{5}\right)&=\frac{\sqrt{5}+1}{4}\\[10pt]
\sin\left(\frac{2\pi}{5}\right)&=\frac{1}{2}\sqrt{\frac{5+\sqrt{5}}{2}}\\[10pt]
\cos\left(\frac{2\pi}{5}\right)&=\frac{\sqrt{5}-1}{4}\\[10pt]
\sin\left(\alpha\pm\beta\right)&=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta\\[10pt]
{\rm{artanh}}\,\theta&=\frac{1}{2}\ln\left(\frac{1+\theta}{1-\theta}\right)\,,\quad\mbox{for}\,|\theta|<1
\end{align}
We can derive the following answer
\begin{equation}
5 \int \frac{\sin(x)}{\sin(5x)} \ dx= \sin\left(\frac{2\pi}{5}\right) \cdot \ln\left\{\frac{\sin\left(x-\frac{2\pi}{5}\right)}{\sin\left(x+\frac{2\pi}{5}\right)}\right\} -\sin\left(\frac{\pi}{5}\right) \cdot \ln\left\{\frac{\sin\left(x-\frac{\pi}{5}\right)}{\sin\left(x+\frac{\pi}{5}\right)}\right\}+C
\end{equation}