But how do we integrate ln|1+tanx| da
X=theta
=int.[ln|cos x|dx] +int.[sin x| dx ] - int.[ln|cos x| dx] ?
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)
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.
Thank you very much for thd explanation and checking the answer
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
Thank you!
To integrate the expression ln|1+tan(x)|, we can break it down into simpler integrals. Let's go step by step:
Step 1: Rewrite the expression ln|1+tan(x)|.
- Recall the identity tan(x) = sin(x)/cos(x).
- Using this identity, we can rewrite the expression as ln|1+sin(x)/cos(x)|.
Step 2: Break down the integral using properties of logarithms.
- We can rewrite the expression ln|1+sin(x)/cos(x)| as ln|cos(x)+sin(x)/cos(x)|.
- Using logarithmic properties, this can be simplified to ln|cos(x)| + ln|1+sin(x)/cos(x)|.
Step 3: Simplify the integral term by term.
- The integral of ln|cos(x)| can be evaluated as follows:
- Recall the integral of ln|x| is x*ln|x| - x + C.
- Substitute x = cos(x) to get the integral of ln|cos(x)| as cos(x) * ln|cos(x)| - cos(x) + C1.
- The integral of ln|1+sin(x)/cos(x)| can be evaluated separately.
Step 4: Evaluate the integral of ln|1+sin(x)/cos(x)|.
- This integral can be solved using the algebraic identity:
- If we have ln|a/b|, we can rewrite it as ln|a| - ln|b|.
- Apply this identity to the integral of ln|1+sin(x)/cos(x)|:
- = ln|1| - ln|cos(x)+sin(x)/cos(x)|.
- The integral of ln|1| is x + C2.
- The integral of ln|cos(x)+sin(x)/cos(x)| can be evaluated further.
Step 5: Simplify and evaluate the final integral.
- Combining the results from steps 3 and 4, we have:
- ln|1+tan(x)| = (cos(x) * ln|cos(x)| - cos(x)) - (x + ln|cos(x)+sin(x)/cos(x)|) + C.
- Simplify the expression further if necessary or perform any additional calculations.
Please note that the symbols C1 and C2 represent the constants of integration, and you may also need to simplify the expression or apply trigonometric identities depending on the specific problem.