First we can simplify:
#csc(x)tan(x) = 1/sin(x)*sin(x)/cos(x) = 1/cos(x) = sec(x)#
so we want to find #int sec(x)dx#. You might have that memorized, but if not the method I use is to multiply by a clever form of 1, #(sec(x)+tan(x))/(sec(x)+tan(x))#:
#int sec(x)dx = int sec(x)*(sec(x)+tan(x))/(sec(x)+tan(x))dx#
#=int (sec^2(x)+sec(x)tan(x))/(sec(x)+tan(x))dx#
Now let #u=sec(x)+tan(x)#, so
#du=(sec(x)tan(x)+sec^2(x))dx#, which matches the numerator of our integrand, so we can substitute and get:
#int 1/u du = ln(|u|)+C#
So the antiderivative we are looking for is: #ln(|sec(x)+tan(x)|)#.
Now we want to evaluate at #x=pi/2# and #x=0# but there's a problem! #sec(x)# is undefined at #x=pi/2#, so we have to do an improper integral:
#int_(0)^(pi/2)sec(x)dx #
#= lim_(b to (pi/2)^-)(ln(|sec(b)+tan(b)|)) - ln(|sec(0)+tan(0)|)#
#= lim_(b to (pi/2)^-)(ln(|sec(b)+tan(b)|)) - ln(|1|)#
#= lim_(b to (pi/2)^-)(ln(|sec(b)+tan(b)|))#
Since #sec(x)\rightarrow oo# as #x\rightarrow pi/2# from the left, and #tan(x)# does the same, the integral diverges because #ln(x)\rightarrow oo# as #x \rightarrow oo#.
