Answer
$-\sqrt{x(1-x)}-\sin^{-1}\sqrt{1-x}+C$
Work Step by Step
Suppose $u=\sqrt {1-x} \implies \dfrac{1}{\sqrt{1-x}}dx=-2\space du$
and $1-x=u^2 \implies x=1-u^2$
Therefore, $I=-2 \int\sqrt{1-u^2} \space du$
or, $I=-2(\dfrac{u}{2}\sqrt{1-u^2}+\dfrac{1}{2}\sin^{-1}u)+C \\=-u\sqrt{1-u^2}-\sin^{-1} (u)+C \\=-\sqrt{x(1-x)}-\sin^{-1}\sqrt{1-x}+C$