Q.Evaluate: $\displaystyle\int e^{\sin^{-1}x}\left(\dfrac{x+\sqrt{1-x^2}}{\sqrt{1-x^2}}\right)dx$

Substitute $t=\sin^{-1}x$ to simplify, reducing it to the standard form $\int e^t(\sin t+\cos t)\,dt$.

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

Let $t=\sin^{-1}x$, so $x=\sin t$, $dx=\cos t\,dt$, and $\sqrt{1-x^2}=\cos t$.

$I=\int e^t\left(\frac{\sin t+\cos t}{\cos t}\right)\cos t\,dt=\int e^t(\sin t+\cos t)\,dt$

Note that $\dfrac{d}{dt}\left(e^t\sin t\right)=e^t\sin t+e^t\cos t=e^t(\sin t+\cos t)$, exactly the integrand.

$I=e^t\sin t+c$

Substituting back $t=\sin^{-1}x$, $\sin t=x$:

$I=x\,e^{\sin^{-1}x}+c$

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