But how do we integrate ln|1+tanx| da

Question

X=theta

=int.[ln|cos x|dx] +int.[sin x| dx ] - int.[ln|cos x| dx] ?

Answer 1

You may not have come across the trick that says that
∫[0,a] f(x) dx = ∫[0,a] f(a-x) dx

It's kind of like integrating from right to left, instead of left to right, but the area is the same.

Now you have to include the limits of integration as part of the substitution.

x = tanθ
when x=0, θ=0
when x=1, θ=π/4
so,
∫[0,1] f(x) dx = ∫[0,π/4] g(θ) dθ

Now we have
A = ∫[0,π/4] ln(1+tanθ) dθ
A = ∫[0,π/4] ln(1+tan(π/4-θ)) dθ
Now, note that

1+tan(π/4-θ) = 1+(tan(π/4)-tanθ))/(1+tan(π/4)tanθ)
= 1+(1-tanθ)/(1+tanθ)
= 2/(1+tanθ)

So, now we have two expressions for A. Add them up and we get

2A = ∫[0,π/4] ln(1+tanθ)+ln(2/(1+tanθ)) dθ
= ∫[0,π/4] ln((1+tanθ)*2/(1+tanθ)) dθ
= ∫[0,π/4] ln(2) dθ
= ln2 θ [0,π/4]
= ln2 (π/4 - 0)
= ln2 π/4

so, A = ln2 π/8

This is one tricky integral!

You can use the same kind of trick with the substitution

x = (1-t)/(1+t)

Answer 2

your idea of splitting the log is wrong, since

ln(1+tanx) = ln((cosx+sinx)/cosx)
= ln(cosx+sinx) - ln(cosx)

but ln(cosx+sinx) ≠ ln(cosx)+ln(sinx)

and you can't integrate those either.

Answer 3

Thank you very much for thd explanation and checking the answer

Answer 4

actually, a bit more examination of your attempt is in order. We have

∫[0,π/4] ln(sinx+cosx) - ln(cosx) dx

recall from your cosine subtraction formula, that this becomes

∫[0,π/4] ln(√2(cos(π/4-x)) - ln(cosx) dx
= ∫[0,π/4] ln(√2)+ln(cos(π/4-x)) - ln(cosx) dx
and with the little trick from above,
= ∫[0,π/4] ln(√2) + ln(cos(x)) - ln(cosx) dx
= ∫[0,π/4] ln(√2) dx
= 1/2 ln(2) * π/4
= ln2 π/8

Answer 5

Thank you!