#### Answer

$$\frac{1}{4}sec4\theta + C$$

#### Work Step by Step

$$\eqalign{
& \int {\sec 4\theta tan4\theta } d\theta \cr
& {\text{use the formula for indefinite integrals of trigonometric functions}} \cr
& \int {\sec ax} \tan axdx = \frac{1}{a}secax + C \cr
& {\text{letting }}a = 4,{\text{ and }}x = \theta ,{\text{ we have}} \cr
& = \frac{1}{4}sec4\theta + C \cr
& {\text{check by differentiation}} \cr
& {\text{ = }}\frac{d}{{d\theta }}\left( {\frac{1}{4}sec4\theta + C} \right) \cr
& {\text{ = }}\frac{d}{{d\theta }}\left( {\frac{1}{4}sec4\theta } \right) + \frac{d}{{d\theta }}\left( C \right) \cr
& = \frac{1}{4}\left( {\sec 4\theta tan4\theta } \right)\left( 4 \right) + 0 \cr
& = \sec 4\theta tan4\theta \cr} $$