Simple indefinate integral

• Sep 15th 2008, 06:01 PM
92stealth
Simple indefinate integral
calc test soon, drawing a lot of blanks pon the subject. I have to find the $integral(5x/sqrt(9-4x^2))$. I came up with an answer, but i'm pretty sure i used some steps that you shouldnt ever use in math....so im stuck. Any help? Thanks....
• Sep 15th 2008, 06:05 PM
o_O
$\int \frac{5x}{\sqrt{9-4x^2}} \ dx$

Do a u-sub: $u = 9 -4x^2 \ \Rightarrow \ du = -8x \ dx \iff x dx = -\frac{du}{8}$

Making the subs:
$= \int \frac{5 \ \frac{du}{8}}{\sqrt{u}} = \frac{5}{8} \int \frac{1}{u^{\frac{1}{2}}} \ du$
• Sep 15th 2008, 06:06 PM
Jhevon
Quote:

Originally Posted by 92stealth
calc test soon, drawing a lot of blanks pon the subject. I have to find the $integral(5x/sqrt(9-4x^2))$. I came up with an answer, but i'm pretty sure i used some steps that you shouldnt ever use in math....so im stuck. Any help? Thanks....

integration by substitution.

let $u = 9 - 4x^2$

or even better (if you can handle it), let $u^2 = 9 - 4x^2$
• Sep 15th 2008, 06:24 PM
92stealth
got it, thank you