I am having some difficulties with the calculation of the following integral. Can somebody help me please?
$\int \frac{dx}{1+a\cos x},\text{ for }0
Thank you in advance
Detailed hint:
Wikipedia calls this the The Weierstrass Substitutiion: when $t=\tan(\theta/2)$, $ \begin{align} \sin(\theta)&=\frac{2t}{1+t^2}\\ \cos(\theta)&=\frac{1-t^2}{1+t^2}\\ \tan(\theta)&=\frac{2t}{1-t^2}\\ \mathrm{d}\theta&=\frac{2\mathrm{d}t}{1+t^2} \end{align} $
Hint 1: $\cos x=\frac{1-t^2}{1+t^2}$ which $t=\tan\frac{x}{2}$.
Hint 2: $t=\tan\frac{x}{2}\Rightarrow dt=\frac{1}{2}\sec^2\frac{x}{2}dx=\frac{1}{2}(t^2+1)dx$.
Substitute, $t = \tan \left(\dfrac x2\right)$. So $x = 2\tan^{-1}t$ and $dx = \dfrac{2dt}{1+t^2}$. And the integral becomes,
$\begin{align*}\int\dfrac{\dfrac{2}{1+t^2}}{1+\dfrac{a(1-t^2)}{1+t^2}}dt &=\dfrac2{(1+a)}\int\dfrac1{1+\left[t\dfrac{\sqrt{1-a}}{\sqrt{1+a}}\right]^2}dt &\color{blue}{u =\left[t\frac{\sqrt{1-a}}{\sqrt{1+a}}\right]\Rightarrow dt = \frac{\sqrt{1+a}}{\sqrt{1-a}}du }\\ &=\dfrac2{(1+a)}\frac{\sqrt{1+a}}{\sqrt{1-a}}\int\dfrac1{1+u^2}du &\color{blue}{\int\dfrac1{1+u^2}du= \tan^{-1}u}\\ \\&=\dfrac2{(1+a)}\frac{\sqrt{1+a}}{\sqrt{1-a}}\tan^{-1}u\\\\ &=\dfrac2{\sqrt{1-a^2}}\tan^{-1}u&\color{blue}{u =\left[t\frac{\sqrt{1-a}}{\sqrt{1+a}}\right]}\\ \\ &=\dfrac{2\tan^{-1}\left[\frac{\sqrt{1-a}}{\sqrt{1+a}}\cdot\tan \left(\dfrac x2\right)\right]}{\sqrt{1-a^2}}\\\\ \end{align*}$
$\displaystyle\Large\therefore \boxed{\int \dfrac1{1+acosx}dx =\dfrac{2\tan^{-1}\left[\frac{\sqrt{1-a}}{\sqrt{1+a}}\cdot\tan \left(\dfrac x2\right)\right]}{\sqrt{1-a^2}} }$