integrate secxdx

Answer 1

recall that if
u = secx, u' = secx tanx

You have

∫ secx dx

Turn that into a fraction by multiplying by

(secx + tanx)

and you have

∫ secx(secx+tanx)/(secx+tanx) dx

Now, if
u = secx+tanx,
du = secx tanx + sec^2 x

and your integral now becomes

∫ du/u = lnu = ln(secx + tanx) + C

tricky, eh?