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: Graph of integrand

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.


$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}



$$ -\int \limits^{2\pi }_{\pi }\frac{\sin \left( x\right) }{\sin \left( x\right) -1} dx= -\int \limits^{2\pi }_{\pi }\frac{1}{\sin \left( x\right) -1} dx-\pi $$

  • $\begingroup$ Could you explain how you came about to this? $\endgroup$ – Will Sherwood Sep 12 '16 at 5:24
  • $\begingroup$ @WillSherwood the integral can be written like $$ -\int \limits^{2\pi }_{\pi }\frac{\sin \left( x\right) -1+1}{\sin \left( x\right) -1} dx=-\int \limits^{2\pi }_{\pi }\frac{1}{\sin \left( x\right) -1} dx-\int \limits^{2\pi }_{\pi }1 dx $$ $\endgroup$ – Refaat M. Sayed Sep 12 '16 at 5:27

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