Respuesta :

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[tex]\displaystyle\int\frac{\cos x}{2+\cos x}\,mathrm dx[/tex]

Let [tex]y=\tan\dfrac x2[/tex], so that

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

Then

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

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