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provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise  18.9 question 10

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Answer: \frac{-1}{\sin ^{-1} x}+C

Hint: Use substitution method to solve this integral.

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


        \begin{aligned} &\text { Let } I=\int \frac{1}{\sqrt{1-x^{2}}\left(\sin ^{-1} x\right)^{2}} d x \\ &\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 \text { then } \end{aligned}

        \begin{aligned} I &=\int \frac{1}{\sqrt{1-x^{2}} \cdot t^{2}} \sqrt{1-x^{2}} d t=\int \frac{1}{t^{2}} d t \\ &=\frac{t^{-2+1}}{-2+1}+c=\frac{t^{-1}}{-1}+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \end{aligned}

            =\frac{-1}{t}+c=\frac{-1}{\sin ^{-1} x}+c\; \; \; \left[\because t=\sin ^{-1} x\right]

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