## University Calculus: Early Transcendentals (3rd Edition)

$$\int\frac{\sqrt x}{\sqrt{1-x}}dx=-\sqrt{x(1-x)}-\sin^{-1}\sqrt{1-x}+C$$ Or (alternative form): $$-\sqrt{x(1-x)}+\sin^{-1}\sqrt{x}+C$$
$$A=\int\frac{\sqrt x}{\sqrt{1-x}}dx$$ We set $u=\sqrt {1-x}$, which means $$du=\frac{(1-x)'}{2\sqrt{1-x}}dx=\frac{-1}{2\sqrt{1-x}}dx$$ $$\frac{1}{\sqrt{1-x}}dx=-2du$$ Also, we can rewrite $1-x=u^2$, so $x=1-u^2$ Therefore, $$A=-2\int\sqrt{1-u^2}du$$ Here apply Formula 45, which states that $$\int\sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C$$ for $a=1$. $$A=-2\Big(\frac{u}{2}\sqrt{1-u^2}+\frac{1}{2}\sin^{-1}u\Big)+C$$ $$A=-u\sqrt{1-u^2}-\sin^{-1}u+C$$ $$A=-\sqrt{1-x}\sqrt{1-(1-x)}-\sin^{-1}\sqrt{1-x}+C$$ $$A=-\sqrt{1-x}\sqrt x-\sin^{-1}\sqrt{1-x}+C$$ $$A=-\sqrt{x(1-x)}-\sin^{-1}\sqrt{1-x}+C$$