Question
Evaluate: $\int \frac{\left\{e^{\sin ^{-1} x}\right\}^2}{\sqrt{1-x^2}} d x$

Answer

Let $I =\int \frac{\left\{e^{\sin ^{-1} \alpha}\right\}^2}{\sqrt{1-x^2}} d x$
Also let sin $x = t$ then, we have
$ d \left(\sin ^{-1} x \right)= dt$
$\Rightarrow \frac{1}{\sqrt{1-x^2}} d x=d t$
$\Rightarrow d x=\sqrt{1-x^2} d t$
Putting $\sin ^{-1} x = t$ and $dx =\sqrt{1-x^2} d t$ in equation $(i),$ we get
$I=\int \frac{\left(e^t\right)^2}{\sqrt{1-x^2}} \times \sqrt{1-x^2} d t$
$=\int e^{2 t} d t$
$=\frac{e^2}{2}+c$
$=\frac{e^{2 \sin ^{-1} x}}{2}+c$
$\therefore I=\frac{\left\{e^{\sin }-1\right\}^2}{2}+c$

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