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Solve the following integral: $$ \frac{2}{\pi}\int_{-\pi}^\pi\frac{\sin\frac{9x}{2}}{\sin\frac{x}{2}}dx $$

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    I tried to convert it into cos + sin – 2017-02-16
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    For any odd $n\in\mathbb{N}$, $$\int_{-\pi}^{\pi}\frac{\sin(nx/2)}{\sin(x/2)}\,dx = \pi.$$ See below. – 2017-02-16
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    Imperative questions are not very well-received here. Please improve your question by adding your attempts / some context. – 2017-02-16

3 Answers 3

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Such integral equals: $$\frac{4}{\pi}\int_{-\pi/2}^{\pi/2}\frac{\sin(9x)}{\sin(x)}\,dx=\frac{4}{\pi}\int_{-\pi/2}^{\pi/2}\frac{e^{9ix}-e^{-9ix}}{e^{ix}-e^{-ix}}\,dx \tag{1}$$ that is: $$ \frac{4}{\pi}\int_{-\pi/2}^{+\pi/2}\left(e^{8ix}+e^{6ix}+\ldots+1+\ldots+e^{-6ix}+e^{-8ix}\right)\,dx=\frac{4}{\pi}\int_{-\pi/2}^{\pi/2}1\,dx=\color{red}{4}\tag{2} $$ since $\int_{-\pi/2}^{\pi/2}\cos(2nx)\,dx = 0$.

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By using $$ \sin A-\sin B=2\cos(\frac{A+B}{2})\sin(\frac{A-B}{2})$$ one has \begin{eqnarray} &&\frac{2}{\pi}\int_{-\pi}^\pi\frac{\sin\frac{9x}{2}}{\sin\frac{x}{2}}dx\\ &=&\frac{2}{\pi}\int_{-\pi}^\pi\frac{\sum_{n=1}^4\bigg[\sin\frac{(2n+1)x}{2}-\sin\frac{(2n-1)x}{2}\bigg]+\sin\frac{x}{2}}{\sin\frac{x}{2}}dx\\ &=&\frac{2}{\pi}\int_{-\pi}^\pi\frac{\sum_{n=1}^42\cos(nx)\sin \frac{x}{2}+\sin\frac{x}{2}}{\sin\frac{x}{2}}dx\\\\ &=&\frac{2}{\pi}\int_{-\pi}^\pi\bigg[2\sum_{n=1}^4\cos(nx)+1\bigg]dx\\ &=&\frac{2}{\pi}\cdot2\pi\\ &=&4. \end{eqnarray} Here $$ \int_{-\pi}^{2\pi}\cos(nx)dx=0$$ is used.

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} {2 \over \pi}\int_{-\pi}^{\pi}{\sin\pars{9x/2} \over \sin\pars{x/2}}\,\dd x & = {2 \over \pi}\oint_{\verts{z} = 1}{\pars{z^{9/2} - z^{-9/2}}/2\ic \over \pars{z^{1/2} - z^{-1/2}}/2\ic}\,{\dd z \over \ic z} = -\,{2\ic \over \pi}\oint_{\verts{z} = 1}{1 \over z^{4}} {z^{9} - 1 \over z - 1}\,\dd z \\[5mm] & = -\,{2\ic \over \pi}\bracks{2\pi\ic\bracks{z^{3}}\pars{z^{9} - 1 \over z - 1}} = 4\,\bracks{z^{3}}\pars{1- z}^{-1} = \bbx{\ds{4}} \end{align}