what is the integration technique to integrate the following function?

Question

What is the integral of the given trigonometric function?

$$\int\dfrac{1}{1-\cos(\alpha)\cos(x)}dx$$

Answer 1

Hint: with $\cos x=\frac12(e^{ix}+e^{-ix})$ and $\cos\alpha=k$ we see $$\int\dfrac{1}{1-\cos(\alpha)\cos(x)}dx=\int\frac{2}{-k}\frac{e^{ix}}{(e^{ix}-\frac1k)^2+(1-\frac{1}{k^2})}dx$$ and using substituation $$e^{ix}-\frac1k=\sqrt{1-\frac{1}{k^2}}\tan\theta$$ we can finish solution.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\int{\dd x \over 1 - \cos\pars{\alpha}\cos\pars{x}} = \int{\dd x \over 1 - \cos\pars{\alpha}\bracks{2\cos^{2}\pars{x/2} - 1}} \\[5mm] = &\ \int{\dd x \over 1 + \cos\pars{\alpha} - 2\cos\pars{\alpha}\cos^{2}\pars{x/2}} = \int{\sec^{2}\pars{x/2} \over \bracks{1 + \cos\pars{\alpha}}\sec^{2}\pars{x/2} - 2\cos\pars{\alpha}}\,\dd x \\[5mm] = &\ \int{\sec^{2}\pars{x/2} \over \bracks{1 + \cos\pars{\alpha}}\tan^{2}\pars{x/2} + 1 - \cos\pars{\alpha}}\,\dd x = \int{\sec^{2}\pars{x/2} \over 2\cos^{2}\pars{\alpha/2}\tan^{2}\pars{x/2} + 2\sin^{2}\pars{\alpha/2}}\,\dd x \\[5mm] = &\ {1 \over 2\sin^{2}\pars{\alpha/2}}\,\tan\pars{\alpha \over 2} \int{\cot\pars{\alpha/2}\sec^{2}\pars{x/2} \over \bracks{\cot\pars{\alpha/2}\tan\pars{x/2}}^{\,2} + 1}\,\dd x \\[5mm] = &\ \bbx{\ds{{2 \over \sin\pars{\alpha}}\, \arctan\pars{\cot\pars{\alpha \over 2}\tan\pars{x \over 2}} + \pars{~\mbox{a constant}~}}} \end{align}