What is #f(x) = int tanx-secx dx# if #f(pi/4)=-1 #?

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Answer 1 · 2018-01-07T13:15:49

#f(x)=-ln|cosx|-ln|secx+tanx|-ln(sqrt2/(sqrt2+1))-1#

As #inttanxdx=-ln|cosx|# and #intsecxdx=ln|secx+tanx|#

#f(x)=int(tanx-secx)dx#

= #-ln|cosx|-ln|secx+tanx|+c#

Hence #f(pi/4)=-ln|1/sqrt2|-ln|sqrt2+1|+c=-1#

or #ln(sqrt2)-ln(sqrt2+1)+c=-1#

or #ln(sqrt2/(sqrt2+1))+c=-1#

or #c=-ln(sqrt2/(sqrt2+1)))-1#

and #f(x)=int(tanx-secx)dx#

= #-ln|cosx|-ln|secx+tanx|-ln(sqrt2/(sqrt2+1))-1#