Question

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

Answer 1

`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`