how to integrate tanu du

Question

it is calculus problem kindly approve that integration equal to ( lin |secu| +c )

Answer 1

∫tan(u)du=∫sin(u)du/cos(u)=-∫(-sin(u)du)cos(u)=-ln|cos(u)|+C=ln|sec(u)|+C.

C is integration constant.

Or:

let x=cos(u), dx=-sin(u)du, so -sin(u)du/cos(u)=-tan(u)du=-dx/x.

-∫dx/x=-ln|x|+C=ln|1/x|+C=ln|1/cos(u)|+C=ln|sec(u)|+C.

If C=ln(A) then the integral can also be written ln(Asec(u)).