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April 16, 2014

April 16, 2014

Posted by **Alex** on Sunday, August 14, 2011 at 3:14am.

sec(x)=ln(secx+tanx)

- calculus -
**Anonymous**, Sunday, August 14, 2011 at 11:51amIf 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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