Question

Evaluate the integral of 1/(x^2-9)^(3/2)dx
Please help!

Answer 1

Umm, use a trig sub.... \(x= 3\sec(\theta)\).

Answer 2

dx=what?

Answer 3

\[
dx =\frac{dx}{d\theta}d\theta= 3\sec(\theta)\tan(\theta) d\theta
\]You don't remember how to do your derivatives?

Answer 4

Its been a long time since I've covered this stuff...Thats why I am here. :P

Answer 5

So how do you plan on being able to replace the original problem with that x and dx?

Answer 6

if you don't mind me mentioning, but I believe if you take a look at Paul's website at http://tutorial.math.lamar.edu/Classes/CalcII/TrigSubstitutions.aspx
He covers the basic trig substitution methods there.

I believe this will give you a better insight of how and especially why to approach such problems in the suggested way. Since you suggested yourself that you're reviewing this stuff, I think this will make things much clearer.

Answer 7

\[
\int \frac{1}{(x^2-9)^{3/2}}dx = \int \frac{3\sec\theta\tan\theta}{((3\sec\theta)^2-9)^{3/2}}d\theta
\]

Answer 8

Okay so what next?

Answer 9

Now I set u= sec(theta)

Answer 10

\[
\sec^2(\theta) - 1 = \tan^2(\theta)
\]

Answer 11

So the bottom is like ((3tan^2(theta)-9)^3/2?

Answer 12

\[
(3\sec\theta)^2-9 = 9\sec^2\theta-9 = 9(\sec^2\theta-1)=9\tan^2\theta = (3\tan\theta)^2
\]

Answer 13

The bottom ends up becoming: \[
3^3|\tan\theta|^3
\]

Answer 14

Okay...Thanks..So as we continue what happens to the top?