Question

Integration of tan^3(2x).sec2x

Answer 1

sec^3 (2x) tan (2x) dx = (1/2) sec^3(2x) tan(2x) d(2x) 
= (1/2) sec^2(2x) sec(2x) tan(2x) dx 
= (1/2) sec^2(2x) d(sec(2x)) 

so 
Integral[sec^3 (2x) tan (2x) dx] = (1/2) Integral[sec^2(2x) d(sec(2x))] 
= (1/2) [sec(2x)]^3 / (3) + constant 
= (1/6) sec^3(2x) + constant 

You might feel more comfortable using the change of variable 
y = sec(2x) 
which would give dy = 2 sec(2x)tan(2x) dx 
so that 
sec^3 (2x) tan (2x) dx = (1/2) y^2 dy 
and then you integrate to get (1/6) y^3 = (1/6) sec^3(2x) + constant