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How do you find integral of ##((secxtanx)/(secx-1))dx##?

Another way of doing this is to consider that ##d(secx) = secxtanx*dx##

That is, the derivative of ##secx## is ##secxtanx##

##=> int(secxtanx)/(secx – 1)dx = int d(secx)/(secx – 1)##

This is the same as letting ##u = secx##

We then have,

##int(du)/(u – 1) = ln(u – 1) + C = ln(secx – 1) + C##


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