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provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise  18.9 question 50

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

Hint: Use substitution method to solve this integral.

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


        \begin{aligned} &\text { Let } I=\int \frac{e^{m \sin ^{-1} x}}{\sqrt{1-x^{2}}} d x \\ &\text { Put } m \sin ^{-1} x=t \Rightarrow m \frac{1}{\sqrt{1-x^{2}}} d x=d t \\ &\Rightarrow \mathrm{d} \mathrm{x}=\frac{\sqrt{1-x^{2}}}{m} \mathrm{dt} \text { then } \end{aligned}

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

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

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

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