Question

What is the antiderivative of sqrt(9-x^2)?

Answer 1

This would most easily be solved using a `Trigonometric Substitution`. Have you learned about that yet? :)

Answer 2

No, I have not.

Answer 3

Hmm is this a homework assignment? :\ I don't think you can solve it without applying a Trig Sub. You might be able to get somewhere with Integration by Parts, but I can't seem to figure it out.

Answer 4

Yeah, It's a homework assignment. Thanks anyways.

Answer 5

Is is possible to solve using U-substitution. Let x = 3sin(u) and dx = 3cos(u) du
\[\int\sqrt{9-x^2}\space dx = \int \sqrt{9-9\sin^2 u}\space dx\]*Note that \(u = \sin^{-1}\left(\dfrac{x}{3}\right)\)
9 - 9sin²u = 9(1 - sin²u) = 9cos²u so \(\displaystyle\int \sqrt{9-9\sin^2 u}\space dx = \int 3\cos(u)\space dx\)
Now subsitive dx to 3cos(u)du
\[\int 3\cos(u)\space dx = 9\int\cos^2(u)\space du\]
We know that \(\cos^2(u) = \dfrac{1}{2}\cos(2u) + \dfrac{1}{2}\)

So \(\displaystyle 9\int \cos^2(u)\space du = 9\int \dfrac{1}{2}\cos(2u) + \dfrac{1}{2}\space du = \dfrac{9}{2}\int \cos (2u) \space du + \dfrac{1}{2}\int du\)

Let s = 2u and ds = 2du \(\Rightarrow \) ds/2 = du: \(\displaystyle \dfrac{9}{4}\int \cos (s) \space ds + \dfrac{1}{2}\int du\)

Answer 6

Can you handle it now?

Answer 7

substitute* not subsitive

Answer 8

@Veillant

Answer 9

I got it, thanks.

Answer 10

Whew! Glad we helped.