Question

Integrate cosx/(2+cosx)

Answer 1

`\displaystyle\int(\cos x)/(2+\cos x)\,mathrm dx`

Let
`y=\tan\frac x2`, so that


`\cos x=\cos2\left(\frac x2\right)=\cos^2\frac x2-\sin^2\frac x2=(1-y^2)/(1+y^2)`

`\mathrm dy=\frac12\sec^2\frac x2\,\mathrm dx\implies2\cos^2\frac x2\,\mathrm dy=\frac2{1+y^2}\,\mathrm dy=\mathrm dx`

Then


`\displaystyle\int(\cos x)/(2+\cos x)\,\mathrm dx=\int((1-y^2)/(1+y^2))/(2+(1-y^2)/(1+y^2))\frac2{1+y^2}\,\mathrm dy`

`=\displaystyle-2\int(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-\frac4{\sqrt3}\arctan\frac y{\sqrt3}+C`

`=2\arctan\left(\tan\frac x2\right)-\frac4{\sqrt3}\arctan\left(\frac1{\sqrt3}\tan\frac x2\right)+C`

`=x-\frac4{\sqrt3}\arctan\left(\frac1{\sqrt3}\tan\frac x2\right)+C`