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?
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?
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.