Evaluate the following integrals:
Assume log(secx + tanx) =t
d(log(secx + tanx)) =dt
(use chain rule to differentiate first differentiate log(secx + tanx) then (secx + tanx)
⇒ =dt
⇒
⇒ secx dx =dt
Put t and dt in the given equation we get
⇒
=
But t = log(secx + tanx)
= ln| log(secx + tanx) | + c.