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Question Number 137173 by JulioCesar last updated on 30/Mar/21

Commented by mathmax by abdo last updated on 31/Mar/21

$$\mathrm{not}\mathrm{defined}\arcsin \mathrm{is}\mathrm{defined}\mathrm{on}[-1,1]\mathrm{but}{\mathrm{x}}^2-1⩽0!$$

Answered by Ñï= last updated on 30/Mar/21

$$I=\int \frac{e^{-a{\sin }^{-1}x}}{i\sqrt{1-x^2}}dx=\frac{1}{i}\int e^{-a{\sin }^{-1}x}d\left({\sin }^{-1}x\right)=\frac{i}{a}{e}^{-a{\sin }^{-1}x}+C$$

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