Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.