Question

integrate tan^3 2xsec2x

Answer 1

in the integral take u = 2x and du = 2dx.
it becomes 1/2*integration ( tan^3(u)sec(u)du )
=1/2*(integration ( tan(u)sec(u)(sec^2(u) - 1) )
substitute s=sec(u) and ds=tan(u)sec(u)
so you get 1/2*(integration ( (s^2 - 1) ds)
=s^3/6 - s/2 + c
=1/6*sec^3(2x) - 1/2*sec(2x) + c.