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

$$\int\sin^{-1}\sqrt xdx=\Big(x-\frac{1}{2}\Big)\sin^{-1}\sqrt x+\frac{1}{2}\sqrt{x(1-x)}+C$$
$$A=\int\sin^{-1}\sqrt xdx$$ We set $u=\sqrt x$, which means $$du=\frac{1}{2\sqrt x}dx=\frac{1}{2u}dx$$ $$dx=2udu$$ Therefore, $$A=\int\sin^{-1}u(2udu)=2\int u\sin^{-1}udu$$ Now apply Formula 106, which states that $$\int x^n\sin^{-1}axdx=\frac{x^{n+1}}{n+1}\sin^{-1}ax-\frac{a}{n+1}\int\frac{x^{n+1}dx}{\sqrt{1-a^2x^2}}$$ for $a=1$ and $n=1$. $$A=2\Big(\frac{u^2}{2}\sin^{-1}u-\frac{1}{2}\int\frac{u^2du}{\sqrt{1-u^2}}\Big)$$ $$A=u^2\sin^{-1}u-\int\frac{u^2du}{\sqrt{1-u^2}}$$ Now apply Formula 49, which states that $$\int\frac{x^2}{\sqrt{a^2-x^2}}dx=\frac{a^2}{2}\sin^{-1}\frac{x}{a}-\frac{1}{2}x\sqrt{a^2-x^2}+C$$ for $a=1$. $$A=u^2\sin^{-1}u-\Big(\frac{1}{2}\sin^{-1}u-\frac{1}{2}u\sqrt{1-u^2}\Big)+C$$ $$A=x\sin^{-1}\sqrt x-\frac{1}{2}\sin^{-1}\sqrt x+\frac{1}{2}\sqrt x\sqrt{1-x}+C$$ $$A=\Big(x-\frac{1}{2}\Big)\sin^{-1}\sqrt x+\frac{1}{2}\sqrt{x(1-x)}+C$$