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प्रश्न
\[\int e^x \cdot \frac{\sqrt{1 - x^2} \sin^{- 1} x + 1}{\sqrt{1 - x^2}} \text{ dx }\]
योग
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उत्तर
\[\text{ Let I }= \int e^x \left[ \frac{\sqrt{1 - x^2} \sin^{- 1} x + 1}{\sqrt{1 - x^2}} \right]dx\]
\[ = \int e^x \left[ \sin^{- 1} x + \frac{1}{\sqrt{1 - x^2}} \right]dx\]
\[\text{ Here}
, f(x) = \sin^{- 1} x\]
\[ \Rightarrow f'(x) = \frac{1}{\sqrt{1 - x^2}}\]
\[\text{ Put e}^x f(x) = t\]
\[ \Rightarrow e^x \sin^{- 1} x = t\]
\[\text{ Diff both sides w . r . t x}\]
\[\left( e^x \sin^{- 1} x + e^x \times \frac{1}{\sqrt{1 - x^2}} \right)dx = dt\]
\[ \therefore I = \int dt\]
\[ = t + C\]
\[ = e^x \sin^{- 1} x + C\]
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