Question

How does the intergal of tan2t equal to (1/2)ln abs(sec2t)+c?

Answer 1

that is the identity, but if you want to know how to get it, you just do
tan = sin/cos
\[\int tan(2t)dt=\int \dfrac{sin(2t)}{cos(2t)}dt\]
Let u = cos (2t) --> du = -2 sin(2t)dt --> sin(2t)dt = -du/2
so, your integral is
\[\int -\dfrac{du}{2u} =-1/2 ln|u| +C \]
and then -1/2 ln |u| = 1/2ln|u|^(-) , replace u = cos(2t) you have
\[1/2ln|(cos(2t)|^{-} =1/2 ln|sec(2t)|\]
therefore, your integral = that + C