Find $\int{x\sqrt{1-x^2}\arcsin{x}dx}$ I tried $$\int{x\sqrt{1-x^2}\arcsin{x}\ \mathrm{d}x}$$$$=\int{\arcsin{x}\ \mathrm{d}\left(\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\right)}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\int{\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\ \mathrm{d}\left(\arcsin{x}\right)}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\int{\frac{2x^3(1-x^2)^\frac{3}{2}}{3\sqrt{1-x^2}}\ \mathrm{d}x}$$$$=\frac{2x^3(1-x^2)^\frac{3}{2}}{3}\arcsin{x}-\frac{2}{3}\int{x^3\ \mathrm{d}x}+\frac{2}{3}\int{x^5\ \mathrm{d}x}$$$$=\cdots$$
This solution isn't true. Where am I wrong?
EDIT: Thank you all for the many different answers! If I could, I'd give you all the accepted answer.

Let $x=\sin{u}$. Substitute $\arcsin{x}=u$ to change variables, notice that $d(\cos^3{u})=-3\sin{u}\cos^2{u}\,du$ and get the integral by parts
$$
\int x\sqrt{1-x^2}\arcsin{x}\,dx =
 \int{u}\,(\sin{u}\cos^2{u}\,du) =
 -\frac{1}{3}\int{u}\,d(\cos^3{u}) =
 \frac{1}{3}\Bigl(\int\cos^3{u}\,du - u\cos^3{u} \Bigr)
$$
After back substitution the last integral looks
$$
\frac{1}{3}\Bigl(\int(1-x^2)\,dx - (1-x^2)^\frac{3}{2}\,\arcsin{x} \Bigr) = 
 \frac{1}{3}\Bigl( x - \frac{x^3}{3} - (1-x^2)^\frac{3}{2}\,\arcsin{x} \Bigr)
$$
Don't forget a constant and get the answer
$$
\int{x\sqrt{1-x^2}\arcsin{x}}\,dx =
 \frac{1}{3}\Bigl( x - \frac{x^3}{3} - (1-x^2)^\frac{3}{2}\,\arcsin{x} \Bigr) + C
$$
Check the math and good luck!
