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.
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.
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 $${1 \over \sin(b - a)} \bigg(\int \tan(x - a)\,dx- \int \tan( x - b)\,dx\bigg)$$ $$ = {1 \over \sin(b - a)} \bigg(-\ln(\cos(x - a)) + \ln(\cos(x - b))\bigg) + C$$