How do you find the integral of ##int sec^2(5x)dx##?

##1/5 tan 5x +C##

Let 5x = u , so that ##dx= 1/5 du##

##int sec^2 5x dx = 1/5intsec^2 u du##

=##1/5 tanu +C##

= ##1/5 tan 5x +C##