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

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.

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    $\begingroup$ No mistake here, $-4$ is a constant and has been thrown in with $C$. $\endgroup$ – Lanier Freeman Feb 19 '16 at 4:41
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There is no difference between $4 \ln(2+ \cos x) - 2\cos x + C$ and $4 \ln(2+ \cos x) - 2\cos x -4 + C$. Either way it means $4 \ln(2+ \cos x) - 2\cos x$ plus some quantity not depending on $x$, which could be any number.

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Nothing's wrong ! Put $u=\cos x +2$

$$-2 (u - 2\ln u) + C=4\ln u -2u +\mathrm C =4 \ln(2+ \cos x) - 2\cos x - 4 + \mathrm C.$$

So $$-4+\textrm C = \rm constant $$

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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

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