How do you find the integral of ##(t)sec^2(2t)dt##?

Use . Remember that ##tantheta=sintheta/costheta##.

Let ##f(t)=t## so that ##f'(t)=1##. Let ##g'(t)=sec^2(2t)## so that ##g(t)=1/2tan(2t)##.

Hence

##inttsec^2(2t)dt##

##=t/2tan(2t)-1/2inttan(2t)dt##

##=t/2tan(2t)-1/2intsin(2t)/cos(2t)dt##

##=t/2tan(2t)+1/4int(-2sin(2t))/cos(2t)dt##

##=t/2tan(2t)+1/4ln(cos(2t))+C##