Posted by Alex on Sunday, August 14, 2011 at 3:14am.
If tan(x/2)=u then
dx=2du/(1+u^2), sec(x)=(1+u^2)/(1-u^2)
Integral of sec(x)dx=Integral of 2du/(1-u^2)=ln((1+u)/(1-u))
(1+tan(x/2))/(1-tan(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)
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