\(\int {sec (3x)tan(3x)} \, dx= \int { \frac{1}{cos(3x)}* \frac{sin(3x)}{cos(3x)} } \, dx\)Substitution:u=cos(3x), du=-3sin(3x)dx, sin(3x)dx= du/-3 Integral becomes:\(\frac{-1}{3} *\int { \frac{1}{ u^{2} } } \, du= \frac{-1}{3} \int {u^{-2} } \, du=\frac{ u^{-1} }{3}= \frac{1}{3u}= \frac{1}{3cos(3x)}+C\)
