1. ## integration

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.

2. Originally Posted by davecs77
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.
$\int u dv = uv - \int v du$

So for
$\int acs(2x) dx$
use
$u = acs(2x)$ ==> $du = -\frac{2}{\sqrt{1 - 4x^2}}$
and
$dv = dx$ ==> $v = x$

Giving:
$\int acs(2x) dx = x~acs(2x) + \int \frac{2x}{\sqrt{1 - 4x^2}}$

For the integral, let $y = 2x$ ==> $dy = 2 dx$ so...
$\int \frac{2x}{\sqrt{1 - 4x^2}} dx = \frac{1}{2} \int \frac{y}{\sqrt{1 - y^2}} dy$

Then $z = 1 - y^2$ ==> $dz = -2y dy$ so
$\int \frac{2x}{\sqrt{1 - 4x^2}} dx = \frac{1}{2} \int \frac{y}{1 - y^2} dy = \frac{1}{2} \cdot -\frac{1}{2} \int \frac{1}{\sqrt{z}} dz$ $= -\frac{1}{4} \int z^{-1/2} dz = -\frac{1}{4} \cdot \frac{1}{\frac{1}{2}} z^{1/2} = -\frac{1}{2} \sqrt{1 - y^2} = -\frac{1}{2} \sqrt{1 - 4x^2}$

Thus
$\int acs(2x) dx = x~acs(2x) + \int \frac{2x}{\sqrt{1 - 4x^2}} = x acs(2x) - -\frac{1}{2} \sqrt{1 - 4x^2}$

-Dan

3. Originally Posted by topsquark
$\int \frac{2x}{\sqrt{1 - 4x^2}}$
Set $\mu=\sqrt{1-4x^2}\implies{d}\mu=-\frac{4x}{\sqrt{1-4x^2}}~dx$, this yields


\begin{aligned}
\int {\frac{{2x}}
{{\sqrt {1 - 4x^2 } }}~dx} &= - \frac{1}
{2}\int {d\mu }\\
&= - \frac{1}
{2}\sqrt {1 - 4x^2 } + k
\end{aligned}

4. Originally Posted by Krizalid
Set $\mu=\sqrt{1-4x^2}\implies{d}\mu=-\frac{4x}{\sqrt{1-4x^2}}~dx$, this yields


\begin{aligned}
\int {\frac{{2x}}
{{\sqrt {1 - 4x^2 } }}~dx} &= - \frac{1}
{2}\int {d\mu }\\
&= - \frac{1}
{2}\sqrt {1 - 4x^2 } + k
\end{aligned}
I know this is shorter, but I prefer smaller steps even if it takes me longer. (It helps my poor mind organize things better!)

-Dan

5. Aha... but you know more stuff than me, so what you said does not count