Question

integration of Tan^32x sec2x dx

Answer 1

Write tan3(2x) as tan2(2x).tan(2x) = [sec2(2x)-1].tan(2x)

Now the function is :

Integral [sec2(2x)-1] . tan(2x) . sec(2x)dx

​Put sec(2x) = u

sec(2x).tan(2x).dx = du/2

Integral [u2-1]/2

= u3/6 - u/4 + C

= sec3(2x)/6 - sec(2x)/4 + C


here is your answer...
hope it helps...