Calculus of a Single Variable, Chapter 8, 8.3, Section 8.3, Problem 26

inttan^5(2x)sec^4(2x)dx
Let's apply integral substitution:u=2x
(du)=2dx
inttan^5(2x)sec^4(2x)dx=inttan^5(u)sec^4(u)(du)/2
Take the constant out and rewrite the integral as,
=1/2intsec^2(u)sec^2(u)tan^5(u)du
Now use the trigonometric identity :sec^2(x)=1+tan^2(x)
=1/2int(1+tan^2(u))sec^2(u)tan^5(u)du
Again apply the integral substitution:v=tan(u)
dv=sec^2(u)du
=1/2int(1+v^2)v^5dv
=1/2int(v^5+v^7)dv
apply the sum rule and power rule,
=1/2(intv^5dv+intv^7dv)
=1/2{(v^(5+1)/(5+1))+(v^(7+1)/(7+1))}
=1/2(v^6/6+v^8/8)
substitute back v=tan(u) and u=2x
=1/2((tan^6(2x))/6+(tan^8(2x))/8)
Add a constant C to the solution,
=1/2(1/6tan^6(2x)+1/8tan^8(2x))+C