What is $f (x) = \int tan x - sec x d x$ $f(x) = int tanx-secx dx$ if $f (\frac{π}{4}) = - 1$ $f(pi/4)=-1$ ?

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Answer 1 · 2018-01-07T13:15:49

$f (x) = - ln | cos x | - ln | sec x + tan x | - ln (\frac{\sqrt{2}}{\sqrt{2} + 1}) - 1$ $f(x)=-ln|cosx|-ln|secx+tanx|-ln(sqrt2/(sqrt2+1))-1$

$f (x) = \int (tan x - sec x) d x$ $f(x)=int(tanx-secx)dx$

= $- ln | cos x | - ln | sec x + tan x | + c$ $-ln|cosx|-ln|secx+tanx|+c$

Hence $f (\frac{π}{4}) = - ln ∣ ∣ ∣ \frac{1}{\sqrt{2}} ∣ ∣ ∣ - ln ∣ ∣ \sqrt{2} + 1 ∣ ∣ + c = - 1$ $f(pi/4)=-ln|1/sqrt2|-ln|sqrt2+1|+c=-1$

or $ln (\sqrt{2}) - ln (\sqrt{2} + 1) + c = - 1$ $ln(sqrt2)-ln(sqrt2+1)+c=-1$

or $ln (\frac{\sqrt{2}}{\sqrt{2} + 1}) + c = - 1$ $ln(sqrt2/(sqrt2+1))+c=-1$

or $c = - ln (\frac{\sqrt{2}}{\sqrt{2} + 1})) - 1$ $c=-ln(sqrt2/(sqrt2+1)))-1$

and $f (x) = \int (tan x - sec x) d x$ $f(x)=int(tanx-secx)dx$

= $- ln | cos x | - ln | sec x + tan x | - ln (\frac{\sqrt{2}}{\sqrt{2} + 1}) - 1$ $-ln|cosx|-ln|secx+tanx|-ln(sqrt2/(sqrt2+1))-1$

What is f(x) = int tanx-secx dxf(x)=∫tanx−secxdxf(x) = int tanx-secx dx if f(pi/4)=-1 f(π4)=−1f(pi/4)=-1 ?