Question

integrate tan^3(2x)sec2x

Answer 1

1. Write tan^3(2x) as tan^2(2x)*tan(2x)
2. Write tan^2(2x) as sec^2(2x)-1 [Using identity tan^2(x)+1=sec^2(x)
3. Let sec2x=t
    Differentiating both sides we get, 2sec(2x)tan(2x)dx = dt
4. Multiply and divide integral by 2

Answer 2

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.