Answer
$$\frac{{{{\tan }^2}5x}}{{10}} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sec }^2}5x} \tan 5xdx \cr
& = \int {\tan 5x{{\sec }^2}5x} dx \cr
& {\text{set }}u = \tan 5x{\text{ then }}\frac{{du}}{{dx}} = 5{\sec ^2}5x,\,\,\,\,\,\,\,\,\frac{{du}}{{5{{\sec }^2}5x}} = dx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\tan 5x{{\sec }^2}5x} dx = \int u {\sec ^2}5x\left( {\frac{{du}}{{5{{\sec }^2}5x}}} \right) \cr
& {\text{cancel the common factor}} \cr
& = \int u \left( {\frac{{du}}{5}} \right) \cr
& = \frac{1}{5}\int u du \cr
& {\text{integrate by using the power rule}} \cr
& = \frac{1}{5}\left( {\frac{{{u^2}}}{2}} \right) + C \cr
& = \frac{{{u^2}}}{{10}} + C \cr
& {\text{write in terms of }}x \cr
& = \frac{{{{\left( {\tan 5x} \right)}^2}}}{{10}} + C \cr
& or \cr
& = \frac{{{{\tan }^2}5x}}{{10}} + C \cr} $$
