arccos(2x)dx
I'm using integration by parts and almost got the right answer
I got xarccos(2x) - (1/2)radical(1-x^2) + c
the radical is messed up I think.
Thanks once again for all the help.
$\displaystyle \int u dv = uv - \int v du$
So for
$\displaystyle \int acs(2x) dx$
use
$\displaystyle u = acs(2x)$ ==> $\displaystyle du = -\frac{2}{\sqrt{1 - 4x^2}}$
and
$\displaystyle dv = dx$ ==> $\displaystyle v = x$
Giving:
$\displaystyle \int acs(2x) dx = x~acs(2x) + \int \frac{2x}{\sqrt{1 - 4x^2}}$
For the integral, let $\displaystyle y = 2x$ ==> $\displaystyle dy = 2 dx$ so...
$\displaystyle \int \frac{2x}{\sqrt{1 - 4x^2}} dx = \frac{1}{2} \int \frac{y}{\sqrt{1 - y^2}} dy$
Then $\displaystyle z = 1 - y^2$ ==> $\displaystyle dz = -2y dy$ so
$\displaystyle \int \frac{2x}{\sqrt{1 - 4x^2}} dx = \frac{1}{2} \int \frac{y}{1 - y^2} dy = \frac{1}{2} \cdot -\frac{1}{2} \int \frac{1}{\sqrt{z}} dz $ $\displaystyle = -\frac{1}{4} \int z^{-1/2} dz = -\frac{1}{4} \cdot \frac{1}{\frac{1}{2}} z^{1/2} = -\frac{1}{2} \sqrt{1 - y^2} = -\frac{1}{2} \sqrt{1 - 4x^2}$
Thus
$\displaystyle \int acs(2x) dx = x~acs(2x) + \int \frac{2x}{\sqrt{1 - 4x^2}} = x acs(2x) - -\frac{1}{2} \sqrt{1 - 4x^2}$
-Dan