# Re: INTEGRATION OF 1/COS(THETA)

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Posted by T.Gracken on May 19, 2001 at 14:39:05:

In Reply to: INTEGRATION OF 1/COS(THETA) posted by Tom B on May 17, 2001 at 13:35:22:

:
: How do I get from:

: y = intgral 1/cos(theta)d(theta)
: or y = integral sec(theta)d(theta)

: to y = ln(sec(theta)+ tan(theta)) + c

: to y = ln tan(pi/4 + theta/2) + c

: thanks

first, I'm going to use t for theta (to simplify my typing).
For the integration, one "trick" that can be used is to multiply the integrand (that's the function you are trying to integrate) by [sec(t)+tan(t)]/[sec(t)+tan(t)].

you now have y =integral of {sec(t)*[sec(t)+tan(t)]/[sec(t)+tan(t)]}dt

now let u=[sec(t)+tan(t)]. then du = sec(t)*[sec(t)+tan(t)]dt.

or y = integral of [1/u]du

which gives y = ln |u| + c

or y = ln |sec(t)+tan(t)|

As for the second part: tan(pi/4 + t/2) = sec(t)+tan(t) is an identity. For hints on how to verify it... use the identity for tan(u+v) and then convert the remaining tangents to sines and cosines. [remember that tan(pi/4)=1] Now, multiply numerator and denominator by [cos(t/2)+sin(t/2)], and finally use the double angle identities.

hope this helps

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