Answer
$$\int \frac{e^{\sin^{-1}\sqrt x}dx}{2\sqrt{x-x^2}}=e^{\sin^{-1}\sqrt x}+C$$
Work Step by Step
$$A=\int \frac{e^{\sin^{-1}\sqrt x}dx}{2\sqrt{x-x^2}}$$
We set $a=\sin^{-1}\sqrt x$, which means $$da=\frac{(\sqrt x)'}{\sqrt{1-(\sqrt x)^2}}dx=\frac{1}{2\sqrt x\sqrt{1-x}}dx$$ $$da=\frac{1}{2\sqrt {x(1-x)}}dx=\frac{dx}{2\sqrt{x-x^2}}$$
Therefore, $$A=\int e^ada$$ $$A=e^a+C$$ $$A=e^{\sin^{-1}\sqrt x}+C$$
