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September 4, 2015
Posted by **Lee** on Saturday, January 7, 2012 at 9:37pm.

integral sec(t/2) dt=

a)ln |sec t +tan t| +C

b)ln |sec (t/2) +tan (t/2)| +C

c)2tan^2 (t/2)+C

d)2ln cos(t/2) +C

e)2ln |sec (t/2)+tan (t/2)| +C

- Calculus -
**drwls**, Sunday, January 8, 2012 at 5:57amintegral sec(t/2) dt =

2*integral sec(t/2) d(t/2)=

2*ln[sec (t/2) + tan (t/2)] +C

- Calculus -
**Lee**, Sunday, January 8, 2012 at 10:36pmThanks