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.