## Calculus (3rd Edition)

$$\int \sec 3\theta \tan 3\theta d\theta =\frac{1}{3} \sec u+c=\frac{1}{3}\sec 3\theta +c$$
Let $u=3\theta$, then $du=3d\theta$ and since we have $(\sec x)'=\sec x\tan x$ then we get $$\int \sec 3\theta \tan 3\theta d\theta =\frac{1}{3}\int \sec u\tan u du=\frac{1}{3} \sec u+c=\frac{1}{3}\sec 3\theta +c$$