If x in [-1, 1],then the value of int e^{sin^ | vedclass

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(D) Let $t = \sin^{-1} x$,then $x = \sin t$ and $dx = \cos t \, dt$.Substituting these into the integral:$\int e^{\sin^{-1} x} \left( \frac{x}{\sqrt{1

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$e^{\sin^{-1} x} \cdot x + c$,where $c$ is the constant of integration.

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(D) Let $t = \sin^{-1} x$,then $x = \sin t$ and $dx = \cos t \, dt$.Substituting these into the integral:$\int e^{\sin^{-1} x} \left( \frac{x}{\sqrt{1-x^2}} + 1 \right) dx = \int e^t \left( \frac{\sin t}{\cos t} + 1 \right) \cos t \, dt$$= \int e^t (\sin t + \cos t) dt$.Using the standard integral formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + c$,where $f(t) = \sin t$ and $f'(t) = \cos t$,we get:$= e^t \sin t + c = e^{\sin^{-1} x} \cdot x + c$.

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

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