How do you find the antiderivative of #e^(2x)*( tan(e^(2x)) )^2#?

1 Answer
Oct 22, 2016

Answer:

#1/2tan(e^(2x))-1/2e^(2x)+C#

Explanation:

#I=inte^(2x)tan^2(e^(2x))dx#

Let #u=e^(2x)#, so that #du=2e^(2x)dx#. Thus:

#I=1/2inttan^2(e^(2x))(2e^(2x)dx)#

#I=1/2inttan^2(u)du#

We can integrate this using the identity #tan^2(u)+1=sec^2(u)#:

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

#I=1/2intsec^2(u)du-1/2intdu#

These are both common integrals:

#I=1/2tan(u)-1/2u+C#

With #u=e^(2x)#:

#I=1/2tan(e^(2x))-1/2e^(2x)+C#