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.
Best Answer
Integration by parts $$\int x \sqrt{1-x^2} \arcsin x\, dx$$ take $$u'(x)=x \sqrt{1-x^2};\;v(x)=\arcsin x$$ so $$u(x)=\int x \sqrt{1-x^2} \, dx=-\frac{1}{2}\int (-2x)(1-x^2)^{-1/2}\, dx=-\frac{1}{3} \left(1-x^2\right)^{3/2}+C$$ and $$v'(x)=\frac{1}{\sqrt{1-x^2}}$$ apply IBP formula $$\int u'(x)v(x)\,dx=u(x)v(x)-\int u(x)v'(x)\,dx$$
$$\int x \sqrt{1-x^2} \arcsin x\, dx=-\frac{1}{3} \left(1-x^2\right)^{3/2}\arcsin x-\int\left(-\frac{1}{3} \left(1-x^2\right)^{3/2}\right)\frac{1}{\sqrt{1-x^2}}\,dx=$$ $$=-\frac{1}{3} \left(1-x^2\right)^{3/2}\arcsin x+\int \frac{1}{3} \left(1-x^2\right)\,dx=$$ $$=-\frac{1}{3} \left(1-x^2\right)^{3/2} \arcsin x+\frac{1}{3} \left(x-\frac{x^3}{3}\right)+C$$ which can be simplified to $$=\frac{1}{9} \left(3x-x^3-3 \left(1-x^2\right)^{3/2} \arcsin x\right)+C$$