Evaluate \int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x.

    \[\begin{array}{l} \int \frac{\mathrm{x} \sin ^{-1} \mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx} \quad\left[\text { Here } 1^{\mathrm{st}} \mathrm{F} \text { unction }=\sin ^{-1} \mathrm{x} \quad 2^{\mathrm{nd}} \mathrm{F} \text { unction }=\frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}}\right] \\ \Rightarrow \text { Using integration by part } 5 \\ \int \text { I.II } \mathrm{dx}=\mathrm{I} \int \mathrm{II} \mathrm{dx}-\int\left(\frac{\mathrm{d} \mathrm{I}}{\mathrm{dx}} \int \mathrm{II} \mathrm{dx}\right) \mathrm{dx} \\ \Rightarrow \int \frac{\mathrm{x} \sin ^{-1} \mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}=\sin ^{-1} \mathrm{x} \int \frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}-\int\left(\frac{1}{\sqrt{1-\mathrm{x}^{2}}} \int \frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}\right) \mathrm{dx} \end{array}\]

Now Solving, Here \int \frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}=\int \frac{-\mathrm{dt}}{2 \sqrt{t}} Putting 1-\mathrm{x}^{2}=\mathrm{t}

    \[\begin{array}{l} -2 \mathrm{x} \mathrm{dx}=\mathrm{dt} \\ \mathrm{x} \mathrm{dx}=\frac{-\mathrm{dt}}{2} \\ =\frac{-1}{2} \frac{\mathrm{t}}{1 / 2}=-\sqrt{\mathrm{t}} \\ =-\sqrt{1-\mathrm{x}^{2}} \end{array}\]

Hence, \int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} \mathrm{dx}=\sin ^{-1} x\left[-\sqrt{1-x^{2}}\right]-\int \frac{1}{\sqrt{1-x^{2}}} \times-\sqrt{1-x^{2}} \mathrm{dx} =-\sqrt{1-x^{2}} \sin ^{-1} x+\frac{x^{2}}{2}+C where C= constant of Integration.