1. ## 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

2. $\displaystyle \int \frac {1}{\sqrt{1-4x^{2}}} \ dx = \int \frac {1}{\sqrt{1-(2x)^{2}}} \ dx$

let u = 2x

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

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

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

let u = 2x

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

$\displaystyle = \frac {1}{2} \arcsin(u) + C = \frac {1}{2} \arcsin(2x) + C$
Thank you, but why is a factor of 1/2 removed form the integration?

4. 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.

5. 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

6. 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 $\displaystyle u = 2x$

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