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

$$\int3\sec^43xdx=\frac{1}{3}\tan3x(\sec^23x+2)+C$$
$$A=\int3\sec^43xdx$$ Use Reduction 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 $a=3$ $$A=3\Big[\frac{\sec^23 x\tan 3 x}{3\times3}+\frac{2}{3}\int\sec^23xdx\Big]$$ $$A=\frac{\sec^23 x\tan 3 x}{3}+2\int\sec^23xdx$$ Apply Formula 99 one more time, for $n=2$ and $a=3$: $$A=\frac{\sec^23 x\tan 3 x}{3}+2\Big(\frac{\sec^03x\tan3x}{3}+\frac{0}{1}\int\sec^03xdx\Big)$$ $$A=\frac{\sec^23 x\tan 3 x}{3}+2\Big(\frac{\tan3x}{3}+0\Big)+C$$ $$A=\frac{\sec^23 x\tan 3 x}{3}+\frac{2\tan3x}{3}+C$$ $$A=\frac{1}{3}\tan3x(\sec^23x+2)+C$$