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