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प्रश्न
` ∫ e^{m sin ^-1 x}/ \sqrt{1-x^2} ` dx
बेरीज
उत्तर
` ∫ e^{m sin ^-1 x}/ \sqrt{1-x^2} ` dx
\[\text{Let} \sin^{- 1} x = t\]
\[ \Rightarrow \frac{1}{\sqrt{1 - x^2}}dx = dt\]
Now,` ∫ e^{m sin ^-1 x}/ \sqrt{1-x^2} ` dx
\[ = \int e^\text{m t} \cdot dt\]
\[ = \frac{e^{mt}}{m} + C\]
` ∫ e^{m sin ^-1 x}/m } ` dx
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