Question

How do you find the antiderivative of `tan^2(x) dx`

Answer 1

You can start by writing `tan^2(x)=sin^2(x)/cos^2(x)` giving:
`inttan^2(x)dx=intsin^2(x)/cos^2(x)dx=`

using: `sin^2(x)=1-cos^2(x)` you get:

`=int(1-cos^2(x))/(cos^2(x))dx=int[1/cos^2(x)-1]dx=`
`=int1/cos^2(x)dx-int1dx=`

`=tan(x)-x+c`

Answer 2

The answer is: `tanx-x+c`.

Remembering that the derivative of `y=tanx` is `y'=1+tan^2x`,

Than:

`inttan^2xdx=int(tan^2x+1-1)dx=`

`=int(tan^2x+1)dx-intdx=tanx-x+c`.