Find $\int x^2\arcsin(2x)dx$ Find $\int x^2\arcsin(2x)dx$
My work.
$\frac{1}{3}\int \arcsin(2x)dx^3=\frac{1}{3}(x^3\arcsin(2x)dx-\int x^3d(\arcsin(2x))$
This yields to finding $\int \frac{2x^3}{\sqrt{(1-4x^2)}}dx$ with which I have problem finding.
$Edit$
$x=\frac{1}{2}sin\theta$
$\frac{2}{8}\int\frac{sin^3\theta cos\theta d\theta}{\sqrt{1-sin^2\theta}} = \frac{2}{8}\int\frac{sin^3\theta cos\theta d\theta}{|cos\theta|}$
Now what should I do with $|cosx|?$It is $cosx$ or $-cosx$.
 A: Using integration by parts:
$\int u\mathrm dv= uv-\int v\mathrm du$
$u=\arcsin {2x} \implies \mathrm du=\frac{2}{\sqrt{1-4x^2}}$
$\,dv=x^2 \mathrm dx \implies v=\frac{x^3}{3}$
$\int x^2 \arcsin{2x}\mathrm dx=\frac{x^3}{3}\arcsin {2x}-\int { \frac{2x^3}{3\sqrt{1-4x^2}}\mathrm dx}=\frac{x^3}{3}\arcsin {2x}+\frac{1}{72}(3\sqrt {1-4x^2} - (1-4x^2)\sqrt {1-4x^2})+c$
$\int { \frac{2x^3}{3\sqrt{1-4x^2}}\mathrm dx}=-\frac{1}{48}\int \frac {1-t}{\sqrt t}\mathrm dt=-\frac{1}{72}(3\sqrt t - t\sqrt t)+c=-\frac{1}{72}(3\sqrt {1-4x^2} - (1-4x^2)\sqrt {1-4x^2})+c$
$t=1-4x^2\implies \mathrm dt= -8x \mathrm dx$
A: I'd like to show you my way to calculate $\int x^2 \arcsin 2x$.
First, let $t=arcsin 2x$. We get $$\int x^2 \arcsin 2x=\int \frac18 \sin^2 t \cdot t \cdot \cos t dt$$.
Use partical integration twice, we get the following equation
$$\int \frac18 \sin^2 t \cdot t \cdot t\cos t dt=\frac{1}{24}t\sin^3t-\int\frac{1}{24}sin^3tdt\\
=\frac{1}{24}t\sin^3t+\frac{1}{24}\cos t\sin^2t-\int\frac{1}{12}\sin t\cos^2 tdt \\
=\frac{1}{24}t\sin^3t+\frac{1}{24}\cos t\sin^2t-\frac{1}{12}\int\sin t+\frac{1}{12}\int \sin^3 tdt$$
from the first equal sign and the last equal sign, we can find
$$-\frac{1}{24}\int \sin^3tdt=\frac{1}{24}\cos t\sin^2t-\frac{1}{12}\int\sin t+\frac{1}{12}\int \sin^3 tdt$$.
So you can get $$\int \sin^3 tdt=\frac13 \cos t\sin^2 t+\frac23\cos t$$
then you can get $\int x^2 \arcsin 2x dx$ by taking $t=\arcsin 2x$ into the equation.
