Let $I = \int x \sin^{-1} x \, dx$ $= \int \sin^{-1} x \cdot x \, dx$ $= \sin^{-1} x \cdot \frac{x^2}{2} - \int \left( \frac{d}{dx} \sin^{-1} x \right) \cdot \frac{x^2}{2} \, dx$ $= \sin^{-1} x \cdot \frac{x^2}{2} - \int \frac{1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} \, dx$ $= \sin^{-1} x \cdot \frac{x^2}{2} - \frac{1}{2} \int \frac{x^2}{\sqrt{1-x^2}} \, dx$ $= \frac{x^2}{2} \sin^{-1} x - \frac{1}{2} I_1$ $I = \frac{x^2}{2} \sin^{-1} x - \frac{1}{2} I_1$ ... (i) where $I_1 = \int \frac{x^2}{\sqrt{1-x^2}} \, dx$Put $x = \sin \theta$ $\Rightarrow dx = \cos \theta \, d\theta$$\therefore I_1 = \int \frac{\sin^2 \theta}{\sqrt{1-\sin^2 \theta}} \cos \theta \, d\theta$$= \int \frac{\sin^2 \theta}{\cos \theta} \cdot \cos \theta \, d\theta$$= \int \sin^2 \theta \, d\theta$$= \frac{1}{2} \int (1 - \cos 2\theta) \, d\theta$$= \frac{1}{2} \int d\theta - \frac{1}{2} \int \cos 2\theta \, d\theta$$= \frac{1}{2} \theta - \frac{1}{2} \frac{\sin 2\theta}{2} + C$$= \frac{1}{2} \theta - \frac{1}{2} \sin \theta \cos \theta + C$$= \frac{1}{2} \sin^{-1} x - \frac{1}{2} x \sqrt{1-x^2} + C$... (ii)From (i) and (ii), we get $\therefore I = \frac{x^2}{2} \sin^{-1} x - \frac{1}{2} \left[ \frac{1}{2} \sin^{-1} x - \frac{1}{2} x \sqrt{1-x^2} \right] + C$$= \frac{1}{4} \sin^{-1} x \cdot (2x^2 - 1) + \frac{x \sqrt{1-x^2}}{4} + C$