Question

\[\int e^x \cdot \frac{\sqrt{1 - x^2} \sin^{- 1} x + 1}{\sqrt{1 - x^2}} \text{ dx }\]

Answer 1

\[\text{ Let I }= \int e^x \left[ \frac{\sqrt{1 - x^2} \sin^{- 1} x + 1}{\sqrt{1 - x^2}} \right]dx\]

\[ = \int e^x \left[ \sin^{- 1} x + \frac{1}{\sqrt{1 - x^2}} \right]dx\]

\[\text{ Here}
, f(x) = \sin^{- 1} x\]

\[ \Rightarrow f'(x) = \frac{1}{\sqrt{1 - x^2}}\]

\[\text{ Put  e}^x f(x) = t\]

\[ \Rightarrow e^x \sin^{- 1} x = t\]

\[\text{ Diff  both  sides  w . r . t x}\]

\[\left( e^x \sin^{- 1} x + e^x \times \frac{1}{\sqrt{1 - x^2}} \right)dx = dt\]

\[ \therefore I = \int dt\]

\[ = t + C\]

\[ = e^x \sin^{- 1} x + C\]