Question

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

Answer 1

`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)`

Answer 2

`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.