# Integral of sqrt(9-x^2)

• Apr 17th 2009, 09:27 PM
lainmaster
Integral of sqrt(9-x^2)
$\displaystyle \int \sqrt {9-x^2} dx$

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?
• Apr 17th 2009, 09:45 PM
TheEmptySet
Quote:

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?

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

Use the substitution $\displaystyle x=3\sin(\theta) \implies dx=3\cos(\theta)d\theta$

$\displaystyle \int \sqrt{9-9\sin^2(\theta)}(3\cos(\theta))d\theta$

$\displaystyle 9\int\\cos(\theta) \sqrt{1-sin^{2}(\theta)}d\theta$

$\displaystyle 9\int \cos^{2}(\theta)d\theta=\frac{9}{2}\int 1+\cos(2\theta)d\theta=\frac{9}{2}\left( \theta + \frac{1}{2}\sin(2\theta)\right)$

$\displaystyle \frac{9}{2}\left( \theta +\sin(\theta)\cos(\theta)\right)=\frac{9}{2}\left( \sin^{-1}\left( \frac{x}{3}\right)+\frac{x}{3}\frac{\sqrt{9-x^2}}{3}\right)=$

$\displaystyle \frac{9}{2}\sin^{-1}\left( \frac{x}{3}\right)+\frac{1}{2}x\sqrt{9-x^2}+c$
• Apr 18th 2009, 05:07 PM
lainmaster
Thanks a lot =)
• Apr 18th 2009, 08:53 PM
derfleurer
If it helps at all, the u' you got when doing integration by parts was incorrect. That'd be why you weren't able to pull off something at least similar.
• Apr 18th 2009, 09:01 PM
lainmaster
Uh... yes, it helps.
To be honest, I never learned how to do derivatives.
Thanks
• Apr 18th 2009, 09:34 PM
derfleurer
Hey, I never learned u-substitution. =p But we get by.