1. ## Indefinite Integral Problem

I would like some help with this indefinite integral problem.

$\displaystyle \int\frac{6dx}{\sqrt{4-(x+1)^2}}$

I know that the answer is going to have arcsin in it, but I don't know what to make u. I've tried $\displaystyle (x+1)$, but I don't know how to get rid of the 4. Any assistance on this problem would be greatly appreciated. Thank you.

2. Originally Posted by xwanderingpoetx
I would like some help with this indefinite integral problem.

$\displaystyle \int\frac{6dx}{\sqrt{4-(x+1)^2}}$

I know that the answer is going to have arcsin in it, but I don't know what to make u. I've tried $\displaystyle (x+1)$, but I don't know how to get rid of the 4. Any assistance on this problem would be greatly appreciated. Thank you.
Note that $\displaystyle \int\frac{6\,dx}{\sqrt{4-(x+1)^2}}=\int\frac{6\,dx}{\sqrt{4\left(1-\left(\tfrac{x+1}{2}\right)^2\right)}}=\int\frac{3 \,dx}{\sqrt{1-\left(\tfrac{x+1}{2}\right)^2}}$

Now apply the substitution $\displaystyle u=\frac{x+1}{2}$. Then you'll see the integral equation for arcsine.

Does this make sense?

3. Yes that makes sense. I didn't realize I could actually pull a 4 out of the $\displaystyle (x+1)^2$. Thank you very much.

4. Originally Posted by Chris L T521
Note that $\displaystyle \int\frac{6\,dx}{\sqrt{4-(x+1)^2}}=\int\frac{6\,dx}{\sqrt{4\left(1-\left(\tfrac{x+1}{2}\right)^2\right)}}=\int\frac{3 \,dx}{\sqrt{1-\left(\tfrac{x+1}{2}\right)^2}}$
Actually, it was done, the only thing you needed to do here, was to put $\displaystyle x+1=2t$ and you have the arcsine.