Answer
$$\frac{1}{5}\sec {x^5} + C$$
Work Step by Step
$$\eqalign{
& \int {{x^4}\sec {x^5}\tan {x^5}} dx \cr
& {\text{set }}u = {x^5}{\text{ then }}\frac{{du}}{{dx}} = 5{x^4},\,\,\,\,\,\,\,\,\frac{{du}}{{5{x^4}}} = dx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {{x^4}\sec {x^5}\tan {x^5}} dx = \int {{x^4}\sec u\tan u} \left( {\frac{{du}}{{5{x^4}}}} \right) \cr
& {\text{cancel the common factor }}{x^4} \cr
& = \int {\sec u\tan u} \left( {\frac{{du}}{5}} \right) \cr
& = \frac{1}{5}\int {\sec u\tan u} du \cr
& {\text{integrate by using the Basic Trigonometric integral }}\int {\sec x\tan x} dx = \sec x + C \cr
& = \frac{1}{5}\sec u + C \cr
& {\text{write in terms of }}x \cr
& = \frac{1}{5}\sec {x^5} + C \cr} $$
