## Thomas' Calculus 13th Edition

$-\sqrt{x(1-x)}-\sin^{-1}\sqrt{1-x}+C$
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, $I=-2(\dfrac{u}{2}\sqrt{1-u^2}+\dfrac{1}{2}\sin^{-1}u)+C \\=-u\sqrt{1-u^2}-\sin^{-1} (u)+C \\=-\sqrt{x(1-x)}-\sin^{-1}\sqrt{1-x}+C$