#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 38 Maths Textbook Solution.

Answer: $\frac{x^{3}}{3}\sin ^{-1}x+\frac{1}{3}\sqrt{1-x^{2}}-\frac{1}{9}\left ( 1-x^{2} \right )^{\frac{3}{2}}+c$

Given:  $\int x^{2}\sin ^{-1}xdx$

Hint: $\int uvdx=u\int vdx-\int \left ( \frac{d}{dx}u\int vdx \right )dx$

Solution:

$I=\int x^{2}\sin ^{-1}x\: dx$

\begin{aligned} &I=\left(\sin ^{-1} x\right) \frac{x^{3}}{3}-\int \frac{1}{3 \sqrt{1-x^{2}}} \times x^{3} d x \ldots \text { applying byparts } \\ &I=\frac{x^{3}}{3} \sin ^{-1} x-\frac{1}{3} \int \frac{x^{2} x}{\sqrt{1-x^{2}}} d x \end{aligned}

Let 1-$x^{2}$=t => -2xdx = dt

\begin{aligned} &=\frac{x^{3}}{3} \sin ^{-1} x+\frac{1}{2} \times \frac{1}{3} \int \frac{(1-t) d t}{\sqrt{t}} \\ &=\frac{x^{3}}{3} \sin ^{-1} x+\frac{1}{6}\left[\int \frac{1}{\sqrt{t}} d t-\int \frac{t}{\sqrt{t}} d t\right] \\ &=\frac{x^{3}}{3} \sin ^{-1} x+\frac{1}{6}\left[2 \sqrt{t}-\frac{2 t^{\frac{3}{2}}}{3}\right] \\ &=\frac{x^{3}}{3} \sin ^{-1} x+\frac{1}{3} \sqrt{1-x^{2}}-\frac{1}{9}\left(1-x^{2}\right)^{\frac{3}{2}}+c \end{aligned}