What is #f(x) = int tanx dx# if #f(pi/8) = 0 #?

Use the definition of #tanx# as the ratio of #sinx# to #cosx#:

#f(x)=inttanxdx=intsinx/cosxdx#

We can put this in the form #int(du)/u=lnabsu+C# if we let #u=cosx#. Differentiating this substitution implies that #du=-sinxdx#, so we need to multiply the integrand by #-1#. Balance this by also multiplying the exterior of the integral by #-1#.

#f(x)=-int(-sinx)/cosxdx=-int(du)/u=-lnabsu+C=-lnabscosx+C#

We can use the initial condition #f(pi/8)=0# to determine the unknown constant of integration #C#:

#0=-lnabscos(pi/8)+C#

#C=ln(cos(pi/8))#

We can find this using a form of the cosine double angle formula: #cos2theta=2cos^2theta-1#. Thus, #cos(pi/4)=2cos^2(pi/8)-1#, and

#cos(pi/8)=sqrt(1/2(1+cos(pi/4)))=sqrt(1/2(1+sqrt2/2))#

#=sqrt(1/2((2+sqrt2)/2))=sqrt(2+sqrt2)/2#

Thus,

#C=ln(sqrt(2+sqrt2)/2)#

and

#f(x)=-lnabscosx+ln(sqrt(2+sqrt2)/2)#

We shall evaluate #f(x)# using a different method.

#inttanxdx = int(tanxsecx)/secxdx#

Let #u = secx# and #du=secxtanx dx#

Then #inttanxdx = int1/udu = lnabsu = lnabs(secx) + "constant"#

#lnsecx -= -lncosx# so the constant of integration will be the same as the other answer.