$\displaystyle \int sec(3x) tan(3x) dx$
It's an easy problem, yet... I don't know how to go about solving it. Can anyone shed some light?
You are trying to find the antiderivative of $\displaystyle \sec{3x}\tan{3x}$
The derivative of $\displaystyle sec{3x}$ is $\displaystyle 3\sec{3x}\tan{3x}$
So the antiderivative of $\displaystyle 3\sec{3x}\tan{3x}$ is $\displaystyle sec{3x}+C$
Then what would be the antiderivative of $\displaystyle \sec{3x}\tan{3x}$?