If f'(x)=tanx, -pi/2<x<pi/2,and f(pi/3)=ln6, the value of f(pi/4) is?

Please show work. I will choose a best answer!

f'(x) = tan(x) , -pi/2 < x < pi/2 , f(pi/3) = ln(6)

First off, if f'(x) = tan(x), it follows that
f(x) = Integral ( tan(x) dx )

We can solve this using substitution, since

f(x) = Integral ( sin(x)/cos(x) dx )
f(x) = Integral ( 1/cos(x) sin(x) dx )

Let u = cos(x). That means
du = -sin(x) dx, so
(-1) du = sin(x) dx

Apply your substitution.

f(x) = Integral ( 1/u (-1) du )
f(x) = (-1) Integral ( (1/u) du )
f(x) = (-1) ln|u| + C

Back-substitute.

f(x) = (-1) ln|cos(x)| + C
f(x) = ln| [cos(x)]^(-1)] | + C
f(x) = ln|sec(x)| + C

Since we are given f(pi/3) = ln(6), we can actually solve for C.
f(pi/3) = ln|sec(pi/3) + C = ln(6)
ln|sec(pi/3)| + C = ln(6)
ln|1/cos(pi/3)| + C = ln(6)
ln|1/(1/2)| + C = ln(6)
ln(2) + C = ln(6)
C = ln(6) - ln(2)
C = ln(6/2)
C = ln(3)

So now we have our function.

f(x) = ln|sec(x)| + ln(6)

And can easily calculate f(pi/4).

f(pi/4) = ln|sec(pi/4)| + ln(6)
f(pi/4) = ln|1/[1/sqrt(2)]| + ln(6)
f(pi/4) = ln|sqrt(2)| + ln(6)
f(pi/4) = ln|6sqrt(2)|

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