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

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

Answers (1)

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}

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