e ^ ^ -1 x [ x+1-x^2 1-x^2 ] d x

Question

$\int e ^{\sin ^{-1 x}}\left[\frac{x+\sqrt{1-x^2}}{\sqrt{1-x^2}}\right] d x$

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Answer

Let $I =\int e ^{\sin ^{-1 x}}\left[\frac{x+\sqrt{1-x^2}}{\sqrt{1-x^2}}\right] d x$
Put $\sin ^{-1} x=t\ldots(i)$
$\therefore x =\sin t$
Differentiating (i) w.r.t. $x$, we get
$\frac{1}{\sqrt{1-x^2}} d x= dt$
$\therefore I =\int e ^{ t }\left[\sin t +\sqrt{1-\sin ^2 t }\right] dt$
$=\int e ^{ t }[\sin t +\cos t ] dt$
Put $f(t)=\sin t$
$\therefore f ^{\prime}( t )=\cos t$
$\therefore I =\int e ^{ t }\left[ f ( t )+ f ^{\prime}( t )\right] dt$
$= e ^{ t } f ( t )+ c$
$= e ^{ t } \sin t + c$
$\therefore I = e ^{\sin ^{-1 x}}(x)+ c$

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