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.

$$\int x\sqrt{1-x^2}sin^{-1}x dx$$
$$x = sin\theta$$
$$cos\theta= \frac{dx}{d\theta}$$
$$dx= cos\theta d\theta$$
$$\int x\sqrt{1-x^2}sin^{-1}x dx= \int sin\theta(\sqrt{cos^2\theta})\theta (cos\theta d\theta)$$
$$=\int \theta sin\theta cos^2\theta d\theta$$
$$=\int \theta sin\theta (1-sin^2\theta) d\theta$$
$$=\int \theta sin\theta -\int \theta sin^3\theta d\theta$$
$$=\int \theta sin\theta -\int \theta\Bigl(\frac{3sin\theta}{4}-\frac{sin3\theta}{4}\Bigl)  d\theta$$
$$=\int \theta sin\theta -\int\frac{3\theta sin\theta}{4}+ \int\frac{\theta sin3\theta}{4}d\theta$$
Applying product rule,
$$\int \theta sin\theta= \theta \int sin\theta - \int(\theta)' \int sin\theta d\theta= sin\theta - \theta cos\theta$$
$$-\int\frac{3\theta sin\theta}{4}= \frac{3\theta cos\theta}{4} -\frac{3 \sin\theta}{4}$$
$$\int\frac{\theta sin3\theta}{4}d\theta=\frac{1}{4}[\frac{sin3\theta}{9}-\frac{\theta cos3\theta}{3}]$$
$$\int x\sqrt{1-x^2}sin^{-1}x dx=sin\theta - \theta cos\theta\ + \frac{3\theta cos\theta}{4} -\frac{3 \sin\theta}{4}+\frac{sin3\theta}{36}-\frac{\theta cos3\theta}{12}$$
$$= \frac{1}{4}(sin \theta - \theta cos\theta)+ \frac{sin3\theta}{36}-\frac{\theta cos3\theta}{12}+ c$$
Substitute theta with corresponding expression in x
