What's the integral of ##int tanx / (cosx)^2 dx##?

##int tanx / (cosx)^2 dx = 1/2 sec^2x +C## (Or, equivalently ##1/2tan^2x +C##, depending on method used.)

Method 1

##tanx/cos^2x = sinx/cosx 1/cos^2x = sinx (cosx)^-3##

Integrate by substitution with ##u=cosx##.

Method 2

##tanx/cos^2x = tanx seec^2x = (secx)(secxtanx)##

Integrate by substitution with ##u=secx##.

Method 3

##tanx/cos^2x = tanx seec^2x##

Integrate by substitution with ##u=tanx##.

This method gets the antiderivative in the form ##1/2tan^2x + C##