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Integrate cosx/(2+cosx)

User UWSkeletor
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1 Answer

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\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
User MadaManu
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