Question

integrate cosx/(2+cosx)

Answer 1

$\displaystyle\int\frac{\cos x}{2+\cos x}\,mathrm dx$

Let $y=\tan\dfrac x2$, so that

$\cos x=\cos2\left(\dfrac x2\right)=\cos^2\dfrac x2-\sin^2\dfrac x2=\dfrac{1-y^2}{1+y^2}$
$\mathrm dy=\dfrac12\sec^2\dfrac x2\,\mathrm dx\implies2\cos^2\dfrac x2\,\mathrm dy=\dfrac2{1+y^2}\,\mathrm dy=\mathrm dx$

Then

$\displaystyle\int\frac{\cos x}{2+\cos x}\,\mathrm dx=\int\frac{\frac{1-y^2}{1+y^2}}{2+\frac{1-y^2}{1+y^2}}\frac2{1+y^2}\,\mathrm dy$
$=\displaystyle-2\int\frac{y^2-1}{(y^2+3)(y^2+1)}\,\mathrm dy$
$=\displaystyle\int\left(\frac2{y^2+1}-\frac4{y^2+3}\right)\,\mathrm dy$
$=2\arctan y-\dfrac4{\sqrt3}\arctan\dfrac y{\sqrt3}+C$
$=2\arctan\left(\tan\dfrac x2\right)-\dfrac4{\sqrt3}\arctan\left(\dfrac1{\sqrt3}\tan\dfrac x2\right)+C$
$=x-\dfrac4{\sqrt3}\arctan\left(\dfrac1{\sqrt3}\tan\dfrac x2\right)+C$