If $x \in[-1,1]$, then the value of $\int \mathrm{e}^{\sin ^{-1} x}\left(\frac{x+\sqrt{1-x^2}}{\sqrt{1-x^2}}\right) \mathrm{d} x$ is

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Put } \sin ^{-1} x=\mathrm{t} \Rightarrow \frac{1}{\sqrt{1-x^2}} \mathrm{~d} x=\mathrm{dt} \ & 	herefore \quad \int \mathrm{e}^{\sin ^{-1} x}\left(\frac{x+\sqrt{1-x^2}}{\sqrt{1-x^2}}ight) \mathrm{d} x=\int \mathrm{e}^{\mathrm{t}}(\sin \mathrm{t}+\cos \mathrm{t}) \mathrm{dt} \ & =e^t \sin \mathrm{t}+\mathrm{c} \ & =x \mathrm{e}^{\sin ^{-1} x}+\mathrm{c} \end{aligned}$

If $x \in[-1,1]$, then the value of $\int \mathrm{e}^{\sin ^{-1}…

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