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

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

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