inverse trig integration

• July 10th 2009, 03:56 PM
slaypullingcat
inverse trig integration
find the integral of dx/sqrt(1-4x^2). I know I should be using sin^-1 indentity when integrating but my problem is making the the coefficient of x equal to 1. I'm doing something wrong with the algebra. Can someone please show me the working. Thank you
• July 10th 2009, 04:32 PM
Random Variable
$\int \frac {1}{\sqrt{1-4x^{2}}} \ dx = \int \frac {1}{\sqrt{1-(2x)^{2}}} \ dx$

let u = 2x

$\frac {1}{2} \int \frac {1}{\sqrt{1-u^{2}}} \ du$

$= \frac {1}{2} \arcsin(u) + C = \frac {1}{2} \arcsin(2x) + C$
• July 10th 2009, 05:18 PM
slaypullingcat
Quote:

Originally Posted by Random Variable
$\int \frac {1}{\sqrt{1-4x^{2}}} \ dx = \int \frac {1}{\sqrt{1-(2x)^{2}}} \ dx$

let u = 2x

$\frac {1}{2} \int \frac {1}{\sqrt{1-u^{2}}} \ du$

$= \frac {1}{2} \arcsin(u) + C = \frac {1}{2} \arcsin(2x) + C$

Thank you, but why is a factor of 1/2 removed form the integration?
• July 10th 2009, 05:25 PM
Random Variable
Quote:

Originally Posted by slaypullingcat
Thank you, but why is a factor of 1/2 removed form the integration?

You can always move a constant outside of the integral sign. You don't have to move it outside, though.
• July 10th 2009, 05:36 PM
slaypullingcat
Quote:

Originally Posted by Random Variable
You can always move a constant outside of the integral sign. You don't have to move it outside, though.

Sorry, I should have been more specific. Where has the half come from? I can't see where it was originally to then move it out the front. Its probably staring me right in the face but I just can't see where it has come from, thank you for your patients
• July 10th 2009, 06:04 PM
Random Variable
Quote:

Originally Posted by slaypullingcat
Sorry, I should have been more specific. Where has the half come from? I can't see where it was originally to then move it out the front. Its probably staring me right in the face but I just can't see where it has come from, thank you for your patients

I made the substitution $u = 2x$

so $du = 2dx$ which implies $dx = \frac {du}{2}$