Originally Posted by

**lainmaster** $\displaystyle \int \sqrt {9-x^2} dx$

I got this in my last exam, and came up with:

$\displaystyle u = \sqrt {9-x^2}$

$\displaystyle u' = \frac{1}{2} (9-x^2)^{-1/2}$

$\displaystyle v' = dx$

$\displaystyle v = x$

So I get

$\displaystyle x \sqrt {9-x^2} -\frac{1}{2} \int x (9-x^2)^{-\frac{1}{2}} dx$

So then I could

$\displaystyle t = 9-x^2$

$\displaystyle -\frac{1}{2} dt = x dx$

to get

$\displaystyle x \sqrt {9-x^2} +\frac{1}{4} \int t^{-\frac{1}{2}} dt$

But the teacher didn't like it. "Wrong choice", she said; and put me a 1.

Her solution was

$\displaystyle 9 \arcsin \frac{x}{3} + \frac{x}{2} \sqrt {9-x^2} + c$

I've been thinking, but can't come up with anything else... and have no idea how to get to the teacher's solution.

By the way, I'm a highschool student, haven't started university yet.

Any help?