calculus
posted by Alex on .
using tangent half substitution show that integral of
sec(x)=ln(secx+tanx)

If tan(x/2)=u then
dx=2du/(1+u^2), sec(x)=(1+u^2)/(1u^2)
Integral of sec(x)dx=Integral of 2du/(1u^2)=ln((1+u)/(1u))
(1+tan(x/2))/(1tan(x/2))=
(cos(x/2)+sin(x/2))/(cos(x/2)sin(x/2))=
Multiply numerator and denominator by
cos(x/2)+sin(x/2)
=(1+sin(x))/cos(x)=sec(x)+tan(x)