What is f(x) = int tanx dxf(x)=∫tanxdx if f(pi/8) = 0 f(π8)=0?

2 Answers
mason m
Aug 30, 2017

f(x)=-lnabscosx+ln(sqrt(2+sqrt2)/2)f(x)=−ln|cosx|+ln(√2+√22)

Explanation:

Use the definition of tanxtanx as the ratio of sinxsinx to cosxcosx:

f(x)=inttanxdx=intsinx/cosxdxf(x)=∫tanxdx=∫sinxcosxdx

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

f(x)=-int(-sinx)/cosxdx=-int(du)/u=-lnabsu+C=-lnabscosx+Cf(x)=−∫−sinxcosxdx=−∫duu=−ln|u|+C=−ln|cosx|+C

We can use the initial condition f(pi/8)=0f(π8)=0 to determine the unknown constant of integration CC:

0=-lnabscos(pi/8)+C0=−ln∣∣cos(π8)∣∣+C

C=ln(cos(pi/8))C=ln(cos(π8))

We can find this using a form of the cosine double angle formula: cos2theta=2cos^2theta-1cos2θ=2cos2θ−1. Thus, cos(pi/4)=2cos^2(pi/8)-1cos(π4)=2cos2(π8)−1, and

cos(pi/8)=sqrt(1/2(1+cos(pi/4)))=sqrt(1/2(1+sqrt2/2))cos(π8)=√12(1+cos(π4))= ⎷12(1+√22)

=sqrt(1/2((2+sqrt2)/2))=sqrt(2+sqrt2)/2= ⎷12(2+√22)=√2+√22

Thus,

C=ln(sqrt(2+sqrt2)/2)C=ln(√2+√22)

and

f(x)=-lnabscosx+ln(sqrt(2+sqrt2)/2)f(x)=−ln|cosx|+ln(√2+√22)

Monzur R.
Aug 30, 2017

f(x) = lnabs(secx) + ln(sqrt(2+sqrt2)/2)f(x)=ln|secx|+ln(√2+√22)

Explanation:

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

inttanxdx = int(tanxsecx)/secxdx∫tanxdx=∫tanxsecxsecxdx

Let u = secxu=secx and du=secxtanx dxdu=secxtanxdx

Then inttanxdx = int1/udu = lnabsu = lnabs(secx) + "constant"∫tanxdx=∫1udu=ln|u|=ln|secx|+constant

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

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