What is the antiderivative of #sec^5xtan^7x#?
1 Answer
Mar 22, 2015
The answer is:
Remembering that:
#sec^2x=1/cos^2x=(sin^2x+cos^2x)/cos^2x=#
-
#intsecxtanxdx=secx+c# ; -
#int[f(x)]^n*f'(x)dx=[f(x)]^(n+1)/(n+1)+c# .
Than: