# Thread: Integral of sqrt(9-x^2)

1. ## 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?

2. 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 $\displaystyle x=3\sin(\theta) \implies dx=3\cos(\theta)d\theta$

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

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

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

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

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

3. Thanks a lot =)

4. 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.

5. Uh... yes, it helps.
To be honest, I never learned how to do derivatives.
Thanks

6. Hey, I never learned u-substitution. =p But we get by.

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# 9-x2 root integration

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