Question:
Evaluate $\int{\frac{\cos x}{2-\cos x}}dx$
My Attempt:
$\int{\frac{1-\tan^2\frac x2}{2(1+\tan^2\frac x2)-(1-\tan^2\frac x2)}}dx$
$=\int{\frac{1-\tan^2\frac x2}{1+3\tan^2\frac x2}}dx$
$=\int{\frac{2-\sec^2\frac x2}{1+3\tan^2\frac x2}}dx$
$=\int{\frac{2}{1+3\tan^2\frac x2}-\frac{\sec^2\frac x2}{1+3\tan^2\frac x2}}dx$
$=I_1-I_2$
$I_2=\frac2{\sqrt3}\tan^{-1}(\sqrt3\tan\frac x2)$
How to solve $I_1?$
$$I=\int{\frac{\cos x}{2-\cos x}dx}\\=-\int{\frac{-\cos x}{2-\cos x}dx}\\=-\int{\frac{2-\cos x-2}{2-\cos x}dx}\\=-\int{1-\frac2{2-\cos x}dx}\\=-x+2\int{\frac{\sec^2\frac x2}{1+3\tan^2\frac x2}dx}\\=-x+\frac4{\sqrt3}\tan^{-1}(\sqrt3\tan\frac x2)+c$$
This matches with the answer given by WolframAlpha