Evaluate the integral 16tan^3xdx?

1 Answer
May 9, 2018

The answer is #=8sec^2x-16ln(|secx|)+C#

Explanation:

The integral is

#I=int16tan^3xdx=16inttan^3xdx#

#=16inttanx(tan^2x)dx#

#=16inttanx(sec^2x-1)dx#

Perform the substitution

#u=secx#, #=>#, #du=secxtanxdx#

Therefore,

#I=16int(u^2-1)((du)/u)#

#=16int(u-1/u)du#

#=16u^2/2-16lnu#

#=8sec^2x-16ln(|secx|)+C#