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explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 8 maths

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Answer: \frac{e^{2 \sin ^{-1} x}}{2}+c

Hint: Use substitution method to solve this integral.

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


        \begin{aligned} &\text { Let } I=\int \frac{\left\{e^{\sin ^{-1} x}\right\}^{2}}{\sqrt{1-x^{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{\left(e^{t}\right)^{2}}{\sqrt{1-x^{2}}} \sqrt{1-x^{2}} d t=\int\left(e^{t}\right)^{2} d t \\ &=\int e^{2 t} d t=\frac{e^{2 t}}{2}+c \quad\left[\because \int e^{a x} d x=\frac{e^{a x}}{a}+c\right] \end{aligned}

            =\frac{e^{2 \sin ^{-1} x}}{2}+c\; \; \; \left[\because t=\sin ^{-1} x\right]

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