Question

$I=\int \frac{dx}{a+b\cos x},$ where $a,b>0anda+b=u,a-b=v$

Answer 1

$I=\int \frac{dx}{a+bcosx}$
Substituting $a+b=u,a-b=v$ and $\frac{x}{2}=t\Rightarrow \frac{dx}{2}=dt$, we get
$I=\int \frac{dt}{u+v{t}^2}$
A) For $v=0$
$I=\int \frac{dt}{u}=\frac{2t}{u}+c=\frac{2}{u}tan\frac{x}{2}+c$
B) For $v>0$
$I=\frac{2}{v}\int \frac{dt}{t^2+u/v}=\frac{2}{v}\sqrt{\frac{v}{u}}{tan}^{-1}\left(t\sqrt{\frac{v}{u}}\right)+c=\frac{2}{\sqrt{uv}}{tan}^{-1}\left(\left(tan\left(\frac{x}{2}\right)\right)\sqrt{\frac{v}{u}}\right)+c$
C) For $v<0$
$I=-\frac{2}{v}\int \frac{dt}{\frac{u}{-v}+t^2}=\frac{2\sqrt{-v}}{-v2\sqrt{u}}log\left|\frac{\sqrt{\frac{u}{-v}}+t}{\sqrt{\frac{u}{-v}}-t}\right|+c=-\frac{1}{\sqrt{-uv}}log\left|\frac{\sqrt{u}+\sqrt{-v}tan\left(x/2\right)}{\sqrt{u}-\sqrt{-v}tan\left(x/2\right)}\right|+c$
D) For $v=-1$, substituting this in (C)
$I=-\frac{1}{\sqrt{u}}log\left|\frac{\sqrt{u}+tan\left(x/2\right)}{\sqrt{u}-tan\left(x/2\right)}\right|+c$