Integrate $\int\frac{\sin(2x)}{2+\cos x} dx$ 
$$\int\frac{\sin(2x)}{2+\cos x} dx $$

Attempt: let $u = 2 + \cos x$. Then $du = -\sin x$. As we can see, $u - 2 = \cos x$, and $\sin(2x) = 2\sin x\cos x$. Using the change of variables method, we see that
$$\int\frac{\sin(2x)}{2+\cos x} dx = -2 \int(u-2)\frac{1}{u}du = -2 \int 1 - 2\frac{1}{u} du = -2 (u - 2\ln u) + C$$
After substituting for $u$, I get $4 \ln(2+ \cos x) - 2\cos x - 4 + C$. According to Wolfram Alpha, the solution is $4 \ln(2+ \cos x) - 2\cos x + C$. Not sure where my mistake is, but any help is greatly appreciated. 

Consider integral 
$$I=\int\frac{\sin(2x)}{2+\cos x}dx.$$ Through the linéarisation formula $\frac{\sin(2x)}{2}=\sin(x)\cos(x),$ we obtain 
$$I=2\int\frac{\sin x\cos x}{2+\cos x}dx,$$ and by the change of variable $u=\cos x, \ x=\arccos x$ and $du=-\sin xdx,$ we arrive at
$$I=-2\int\frac{udu}{2+u}du=-2\int\frac{u+2-2}{2+u}du$$ 
$$=-2\left(\int du-2\int\frac{1}{2+u}du\right)=-2u+4\ln|u+2|+C$$ which gives the result by replacing $u$ by $\cos x$. Thanks
