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∫ E X ⋅ √ 1 − X 2 Sin − 1 X + 1 √ 1 − X 2 D X - Mathematics

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प्रश्न

\[\int e^x \cdot \frac{\sqrt{1 - x^2} \sin^{- 1} x + 1}{\sqrt{1 - x^2}} \text{ dx }\]
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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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अध्याय 19: Indefinite Integrals - Exercise 19.26 [पृष्ठ १४३]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.26 | Q 18 | पृष्ठ १४३

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