# Evaluate: $\int \sec (x-a) \sec (x-b)\; dx$?

How to do this integration: $$\int \sec (x-a) \sec (x-b)\ dx?$$

I want to this in the shortest possible way. Please guide me through.

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Hint: try using $\sec(y) = 1/\cos(y)$ and $\cos(y)\cos(z) = \frac{1}{2}[\cos(y+z)+\cos(y-z)]$ –  Dilip Sarwate Dec 8 '11 at 16:14
By "solve" I guess you mean "evaluate"... –  GEdgar Dec 8 '11 at 16:19
Have you made any progress by using Dilip's hint? –  The Chaz 2.0 Dec 8 '11 at 16:32
Probably not the shortest possible way but you can try to: 1. Type "integrate 1/[cos(x-a)*cos(x-b)]" into wolframalpha; 2. wait for a while; 3. click on "show steps" wolframalpha.com/input/… –  Martin Sleziak Dec 8 '11 at 18:07

This can be done via the following trick. Rewrite your integral as $${1 \over \sin(b - a)} \int {\sin(b - a) \over \cos(x - a)\cos(x - b)}\,dx$$ Note that $b - a = (x - a) - (x -b)$, so by the sine subtraction formula you have $$\sin(b - a) = \sin(x - a)\cos(x - b) - \sin(x - b)\cos(x - a)$$ Subtituting this back into your integral it becomes $${1 \over \sin(b - a)} \bigg(\int \tan(x - a)\,dx- \int \tan( x - b)\,dx\bigg)$$ $$= {1 \over \sin(b - a)} \bigg(-\ln(\cos(x - a)) + \ln(\cos(x - b))\bigg) + C$$