# Simplification of an integral please explain

1. Apr 19, 2010

### phys-lexic

In doing math I try not to memorize shortcuts/simplifications, but instead understand what's happening. When studying integrals, my professor gave out a few "simplifications" for us to use on problems; most of which I have been able to figure out, except one. I just cannot seem to figure out the relationship given, please help clarify/explain. Thankyou.

$$\int(\sqrt{a^2-u^2})du$$ = $$\left(\frac{u}{2}\right)$$$$\times$$$$\left(\sqrt{a^2-u^2}\right)$$ + $$\left(\frac{a^2}{2}\right)$$$$\times$$$$\left(sin^{-1}\left(\frac{u}{a}\right)\right)$$ + C

I have tried:
- u substitution
- trig substitution
- IBP

*It could be my steps, maybe I'm just doing the intermediates wrong.
**it took a really long time to put that formula in

2. Apr 19, 2010

### tiny-tim

Hi phys-lexic!

(have a square-root: √ and try using the X2 tag just above the Reply box )

Try substituting u = asinθ, then use a couple of the standard trigonometric identities

Last edited: Apr 19, 2010
3. Apr 19, 2010

### g_edgar

What he said. An integral with a difference of squares suggests a certain right triangle, and that can be used to suggest a certain trigonometric substitution. In this case, $u = a \sin\theta$

4. Apr 19, 2010

### phys-lexic

Thankyou, the trig-substitution worked and I set up the correct proof. If you wouldn't mind, we were assigned homework for an exam review. I finished the regular problems, but he gave us three "critical thinking" problems that would be way harder than the exam. I can do them all, but one is giving me issue (I cannot seem to find the final value of x after doing the integral). Here is a link to the problem I posted in the homework section, thankyou again).