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Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 57 maths textbook solution

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Answer: \frac{1}{4}\left(\sin ^{-1} x\right)^{4}+c

Hint: Use substitution method to solve this integral

Given:  \int \frac{\left(\sin ^{-1} x\right)^{3}}{\sqrt{1-x^{2}}} d x


        \text { Let } I=\int \frac{\left(\sin ^{-1} x\right)^{3}}{\sqrt{1-x^{2}}} d x

        \begin{aligned} &\text { put } \sin ^{-1} x=t \Rightarrow \frac{1}{\sqrt{1-x^{2}}} d x=d t \\ &\Rightarrow d x=\sqrt{1-x^{2}} d t \end{aligned}

        I=\int \frac{t^{3}}{\sqrt{1-x^{2}}} \cdot \sqrt{1-x^{2}} d t=\int t^{3} d t

            =\frac{t^{3+1}}{3+1}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right]

            =\frac{1}{4} t^{4}+c=\frac{1}{4}\left(\sin ^{-1} x\right)^{4}+c \quad\left[\because t=\sin ^{-1} x\right]


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