Integral $\lim_{k\rightarrow\infty}\int_\pi^{2\pi}{\sum_{n=1}^k}\left(\sin x\right)^n\ \text{dx}$ I need help with this integral:
$$I=\lim_{k\rightarrow\infty}\int_\pi^{2\pi}{\sum_{n=1}^k}\left(\sin x\right)^n\ \text{dx.}$$
The integrand graph looks like this:

I have rewritten the integral as $$I=-\int_\pi^{2\pi}\frac{\sin x}{\sin x-1}\ \text{dx}$$
But I'm not sure how to move forwards from here.
 A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
I & = \color{#f00}{-\int_{\pi}^{2\pi}
{\sin\pars{x} \over \sin\pars{x} - 1}\,\dd x} =
\int_{\pi}^{2\pi}{\sin\pars{x} + \sin^{2}\pars{x} \over \cos^{2}\pars{x}}\,\dd x
\\[5mm] & =
\int_{\pi}^{2\pi}{\sin\pars{x} \over \cos^{2}\pars{x}}\,\dd x
+ \int_{\pi}^{2\pi}\tan^{2}\pars{x}\,\dd x =
\int_{\pi}^{2\pi}{\sin\pars{x} \over \cos^{2}\pars{x}}\,\dd x
+ \int_{\pi}^{2\pi}\sec^{2}\pars{x}\,\dd x - \int_{\pi}^{2\pi}\dd x
\\[5mm] &=
\left.{1 \over \cos\pars{x}} + \tan\pars{x} - x\,\right\vert_{\ \pi}^{\ 2\pi} =
\color{#f00}{2 - \pi}
\end{align}
A: Hint
$$ -\int \limits^{2\pi }_{\pi }\frac{\sin \left( x\right) }{\sin \left( x\right) -1} dx= -\int \limits^{2\pi }_{\pi }\frac{1}{\sin \left( x\right) -1} dx-\pi $$
