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

$$\int^{\sqrt3/2}_0\frac{dy}{(1-y^2)^{5/2}}=2\sqrt3$$
$$A=\int^{\sqrt3/2}_0\frac{dy}{(1-y^2)^{5/2}}$$ Set $\sin u=y$, which means $$dy=\cos udu$$ For $y=0$, $u=0$ and for $y=\sqrt3/2$, $u=\pi/3$ We would have $$A=\int^{\pi/3}_0\frac{\cos udu}{(1-\sin^2u)^{5/2}}$$ $$A=\int^{\pi/3}_0\frac{\cos udu}{(\cos^2u)^{5/2}}=\int^{\pi/3}_0\frac{\cos udu}{\cos^5u}=\int^{\pi/3}_0\frac{du}{\cos^4u}$$ $$A=\int^{\pi/3}_0\sec^4udu$$ Now apply Formula 99, which states that $$\int \sec^naxdx=\frac{\sec^{n-2}ax\tan ax}{a(n-1)}+\frac{n-2}{n-1}\int\sec^{n-2}axdx$$ for $n=4$ and $a=1$ $$A=\frac{\sec^2u\tan u}{3}\Big]^{\pi/3}_0+\frac{2}{3}\int^{\pi/3}_0\sec^2udu$$ $$A=\frac{1}{3}(\sec^2(\pi/3)\tan(\pi/3)-0)+\frac{2}{3}\tan u\Big]^{\pi/3}_0$$ $$A=\frac{1}{3}(4\sqrt3)+\frac{2}{3}(\sqrt3-0)$$ $$A=\frac{4\sqrt3}{3}+\frac{2\sqrt3}{3}=2\sqrt3$$