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

2 Answers
Aug 30, 2017

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

Explanation:

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)

Aug 30, 2017

f(x) = lnabs(secx) + ln(sqrt(2+sqrt2)/2)

Explanation:

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.