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What is the integral of the following: $$\frac{\sin 2x}{1+\sin x}$$

I know that $\sin 2x = 2\sin x \cos x$ and then I substituted $u$ for $\sin x$ but then I got stuck after that. Can you help?

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We have $du=\cos x\,dx$. Replace $\cos x\,dx$ by $du$, and $\sin x$ by $u$. So we want $$\int 2\frac{u}{1+u}\,du.$$ Now note that $\dfrac{u}{1+u}=1-\dfrac{1}{1+u}$. The integral is $2u-2\log(|1+u|)+C$.

It is somewhat easier to make the substitution $v=1+\sin x$. Then $dv=\cos x\,dx$, and $\sin x=v-1$. When you subsitute you get $$\int 2\frac{v-1}{v}\,du.$$

But $\dfrac{v-1}{v}=1-\dfrac{1}{v}$. Now the integration is easy.

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Okay, I got the integral of (2*u*du)/(1+u). I'm not sure where to go from there to get what you have. – Lizi Dec 11 '12 at 4:29
I missed the divides sign, see revised answer. – André Nicolas Dec 11 '12 at 4:32

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