# Integration of inverse trig functions

1. Feb 1, 2014

### QuantumCurt

This is for Calculus II. I've found most of the integrations on inverse trig functions to be pretty simple, but for some reason this one is throwing me off.

1. The problem statement, all variables and given/known data

$$\int\frac{x+5}{\sqrt{9-(x-3)^2}}dx$$

3. The attempt at a solution

I started by breaking the integral up into two separate integrals be rewriting the numerator-

$$\int\frac{x-3}{\sqrt{9-(x-3)^2}}dx+\int\frac{8}{\sqrt{9-(x-3)^2}}dx$$

The second term integrates easily to an arcsin-

$$\int\frac{x-3}{\sqrt{9-(x-3)^2}}dx+8arcsin(\frac{x-3}{3})+c$$

I feel like I'm missing something on the first term though. I'm trying to integrate it by completing the du, and it seems wrong. I'm using the inside of the radical as my u.

$$u=9-(x-3)^2$$
$$du=-2(x-3)$$

Then I supplied a -1/2 to offset the constant-

$$\frac{-1}{2}\int\frac{-2(x-3)}{\sqrt{9-(x-3)^2}}dx+8arcsin(\frac{x-3}{3})+c$$

I tried integrating the first term to a natural logarithm at first, which was incorrect. I can't integrate it to an arcsin because of the variable in the numerator. Can I move the denominator to the top and rewrite it as a -1/2 power?

$$\frac{-1}{2}\int[(9-(x-3)^2)]^{-1/2}[-2(x-3)dx]+8arcsin(\frac{x-3}{3})+c$$

Giving me-

$$-\sqrt{9-(x-3)^2}+8arcsin(\frac{x-3}{3})+c$$

$$-\sqrt{6x-x^2}+8arcsin(\frac{x-3}{3})+c$$

Does that look right? For some reason it just doesn't seem right to me. Any help would be much appreciated.

2. Feb 1, 2014

3. Feb 1, 2014

### Dick

Yes, that looks right.

4. Feb 1, 2014

### QuantumCurt

I feel like smacking my head into the desk. I have no idea why I didn't think to do that.

I guess when you're studying for hours and hours straight, you sometimes forget to do the obvious...lol

I differentiated it, and it is indeed right. Thanks.