\(\int(a+b\cos x)^{-2}\,dx\)

[jacks]

\(\displaystyle \int \frac{1}{(a+b\cos x)^2}dx\), where \(a,b >0\)

[Grigorios Kostakos]

I guess there are smarter solutions than the one that follows and is based on standard methods of integration:


We assume that \(b>a>0\). The case \(b>a>0\) can be treated analogously. The case \(b=a>0\) is easy.
By the substitution \(t=\tan{\tfrac{x}{2}}\) we have \(\cos{x}={\displaystyle{\frac{1-t^2}{1+t^2}}}\) and \({\displaystyle{\frac{2}{1+t^2}}}\,dt=dx\). So
\begin{align*}
\displaystyle \int{\frac{1}{(a+b\,\cos x)^2}\,dx}=& \int{\frac{1}{\bigl(a+b\,\frac{1-t^2}{1+t^2}\bigr)^2}\,\frac{2}{1+t^2}\,dt}=\int{\frac{2+2t^2}{\bigl({a+b-(b-a)\,t^2}\bigr)^2}\,dt}\\
\mathop{=\!=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c}
{u\,=\,t\,\sqrt{b-a}}\\
{\frac{1}{\sqrt{b-a}}\,du\,=\,dt} \\
\end{subarray}}&\,\frac{1}{\sqrt{b-a}}\int{\frac{2+\frac{2u^2}{b-a}}{\bigl({a+b-u^2}\bigr)^2}\,du}=\frac{1}{({b-a})^{\frac{3}{2}}}\int{\frac{2\,(b-a)+2u^2}{\bigl({u-\sqrt{a+b}\,}\bigr)^2\bigl({u+\sqrt{a+b}\,}\bigr)^2}\,du}\\
=&\, \frac{1}{({b-a})^{\frac{3}{2}}}\int{\frac {a}{(a+b)^{\frac{3}{2}}\left({u-\sqrt {a+b}\,}\right)}\,du}+\frac{1}{({b-a})^{\frac{3}{2}}}\int{\frac {b}{(a+b)\left({u-\sqrt {a+b}\,}\right)^{2}}\,du}\,-\\
& \frac{1}{({b-a})^{\frac{3}{2}}}\int{\frac{a}{(a+b)^{\frac{3}{2}} \left({u+ \sqrt {a+b}\,}\right)}\,du}+\frac{1}{({b-a})^{\frac{3}{2}}}\int{\frac{b}{( a+b)\left({u+\sqrt{a+b}\,}\right)^{2}}\,du}\\
=&\, \frac{a}{({b-a})^{\frac{3}{2}}(a+b)^{\frac{3}{2}}}\,\log\bigl|{u-\sqrt {a+b}\,}\bigr|-\frac {b}{({b-a})^{\frac{3}{2}}(a+b)\left({u-\sqrt{a+b}\,}\right)}\,-\\
&\frac{a}{({b-a})^{\frac{3}{2}}(a+b)^{\frac{3}{2}}}\,\log\bigl|{u+\sqrt{a+b}\,}\bigr|-\frac{b}{({b-a})^{\frac{3}{2}}( a+b)\left({u+\sqrt{a+b}\,}\right)}+c\\
=& \,\frac{2b\,u}{(a+b)\,(b-a)^{\frac{3}{2}}\,(a+b-u^2)}+\frac{a}{(a+b)^{\frac{3}{2}}\,(b-a)^{\frac{3}{2}}}\,\log\Bigl|{\tfrac{u-\sqrt{a+b}}{u+\sqrt{a+b}}\,}\Bigr|+c\\
\stackrel{u\,=\,\sqrt{b-a}\,\tan{\frac{x}{2}}}{=\!=\!=\!=\!=\!=\!=\!=\!=}&\,\frac{2b\,\sqrt{b-a}\,\tan{\frac{x}{2}}}{(a+b)\,(b-a)^{\frac{3}{2}}\,\bigl({a+b-(b-a)\,\tan^2{\frac{x}{2}}}\bigr)}+\frac{a}{(a+b)^{\frac{3}{2}}\,(b-a)^{\frac{3}{2}}}\,\log\Bigl|{\tfrac{\sqrt{b-a}\,\tan{\frac{x}{2}}-\sqrt{a+b}}{\sqrt{b-a}\,\tan{\frac{x}{2}}+\sqrt{a+b}}\,}\Bigr|+c\\
=& \,\frac{2b\,\tan{\frac{x}{2}}}{(b^2-a^2)\,\bigl({a+b-(b-a)\,\tan^2{\frac{x}{2}}}\bigr)}+\frac{a}{(b^2-a^2)^{\frac{3}{2}}}\,\log\Bigl|{\tfrac{\sqrt{b-a}\,\tan{\frac{x}{2}}-\sqrt{a+b}}{\sqrt{b-a}\,\tan{\frac{x}{2}}+\sqrt{a+b}}\,}\Bigr|+c\,.\quad\square
\end{align*}