Calculus of a Single Variable, Chapter 8, 8.3, Section 8.3, Problem 25

inttan^3(2t)sec^3(2t)dt
Apply integral substitution: u=2t
du=2dt
inttan^3(2t)sec^3(2t)dt=inttan^3(u)sec^3(u)(du)/2
Take the constant out,
=1/2inttan^3(u)sec^3(u)du
Rewrite the integral as,
=1/2intsec^3(u)tan^2(u)tan(u)du
Now use the trigonometric identity:tan^2(x)=sec^2(x)-1
=1/2intsec^3(u)(sec^2(u)-1)tan(u)du
Again apply the integral substitution:v=sec(u)
dv=sec(u)tan(u)du
=1/2intv^2(v^2-1)dv
=1/2int(v^4-v^2)dv
Apply the sum and power rule,
=1/2(intv^4dv-intv^2dv)
=1/2{(v^(4+1)/(4+1))-(v^(2+1)/(2+1))}
=1/2(v^5/5-v^3/3)
Substitute back v=sec(u) and u=2t, and add a constant C to the solution,
=1/2((sec^5(2t))/5-(sec^3(2t))/3)+C