Answer
$$\frac{{{{\tan }^{11}}4x}}{{44}} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\tan }^{10}}4x} {\sec ^2}4xdx \cr
& {\text{set }}u = 4x,\,\,\,\,du = 4dx,\,\,\,\,\,\frac{1}{4}du = dx \cr
& {\text{use the substitution}} \cr
& \int {{{\tan }^{10}}4x} {\sec ^2}4xdx = \int {{{\tan }^{10}}u} {\sec ^2}u\left( {\frac{1}{4}du} \right) \cr
& = \frac{1}{4}\int {{{\tan }^{10}}u} {\sec ^2}udu \cr
& \cr
& {\text{use the substitution }}v = \tan u,\,\,\,\,\,\,\,\,dv = {\sec ^2}udu \cr
& \frac{1}{4}\int {{{\tan }^{10}}u} {\sec ^2}udu = \frac{1}{4}\int {{v^{10}}} dv \cr
& {\text{integrate}} \cr
& = \frac{1}{4}\left( {\frac{{{v^{11}}}}{{11}}} \right) + C \cr
& = \frac{{{v^{11}}}}{{44}} + C \cr
& {\text{replace }}v = \tan u \cr
& = \frac{{{{\tan }^{11}}u}}{{44}} + C \cr
& {\text{replace }}u = 4x \cr
& = \frac{{{{\tan }^{11}}4x}}{{44}} + C \cr} $$
