How do you evaluate the integral #int tan theta d(theta)# from 0 to #pi/2#?

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Answer 1 · 2016-07-07T07:22:30

integral non convergent :-(

#int_0^(pi/2) d theta tan theta#

#= int_0^(pi/2) d theta sin theta / cos theta#

and because #d/dx ln f(x) = 1/f(x) f'(x)#

#= - [ln cos theta ]_0^(pi/2)#

#= [ln cos theta ]_(pi/2)^0#

#= [ln cos 0 ] - [ln cos( pi/2) ]#

#= [ln 1 ] - [color{red}{ln 0} ]#