1. ## Integral help

Is this a simple negative exponent, power chain rule

Or an inverse sin integral?

And if it's an inverse sin integral, how do i get rid of the first x under the radical?

$\int \frac{dx}{\sqrt{3x-x^2}}$

2. Originally Posted by penguinpwn
Is this a simple negative exponent, power chain rule

Or an inverse sin integral?

And if it's an inverse sin integral, how do i get rid of the first x under the radical?

$\int \frac{dx}{\sqrt{3x-x^2}}$
it's a pain-in-the-### arcsin ...

$3x - x^2 =$

$-\left(x^2 - 3x + \frac{9}{4} - \frac{9}{4}\right) =
$

$-\left[\left(x - \frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2\right] =$

$\left(\frac{3}{2}\right)^2 - \left(x - \frac{3}{2}\right)^2$

now use the general form of the integral ...

$\int \frac{du}{\sqrt{a^2 - u^2}} = \arcsin\left(\frac{u}{a}\right) + C$

3. Originally Posted by skeeter
it's a pain-in-the-### arcsin ...

$3x - x^2 =$

$-\left(x^2 - 3x + \frac{9}{4} - \frac{9}{4}\right) =
$

$-\left[\left(x - \frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2\right] =$

$\left(\frac{3}{2}\right)^2 - \left(x - \frac{3}{2}\right)^2$

now use the general form of the integral ...

$\int \frac{du}{\sqrt{a^2 - u^2}} = \arcsin\left(\frac{u}{a}\right) + C$
Where did the 9/4 came from?

4. Originally Posted by penguinpwn
Where did the 9/4 came from?
ever heard of completing the square?

5. Originally Posted by skeeter
ever heard of completing the square?
Ooooohhhhhhh, haha I see it now.

Thanks for the help!