oh no im sorry i mean triangle trig, where its like du/sqrt 1+u^2 and u make a triangle with the sqrt on the hyp, u on the tall side and 1 on the base. then find that sec theta = sqrt 1+u^2 tan theta = u sec ^2 theta dtheta = du
We have 1+u^2 and 1+tan^2=sec^2. By making that substitution, we have a well known trig identity. When we make u=tan and substitute, we have 1+(tan)^2=1+tan^2=sec^2. Then by taking the square root, we get sec.