Mathematics Asked on January 7, 2022
My work so far
$${displaystyleint_0^1}dfrac{mathrm{e}^{12x}-mathrm{e}^{-12x}}{mathrm{e}^{12x}+mathrm{e}^{-12x}},mathrm{d}x$$
using substitution with $u=mathrm{e}^{12x}+mathrm{e}^{-12x}$
$$={displaystyleint}dfrac{1}{12u},mathrm{d}u$$
$$=class{steps-node}{cssId{steps-node-1}{dfrac{1}{12}}}{displaystyleint}dfrac{1}{u},mathrm{d}u$$
$${displaystyleint}dfrac{1}{u},mathrm{d}u$$
which is the standard integral:
$$=lnleft(uright)$$
$$=dfrac{lnleft(uright)}{12}$$
and using substitution with $u=mathrm{e}^{12x}+mathrm{e}^{-12x}$
$$=dfrac{lnleft(mathrm{e}^{12x}+mathrm{e}^{-12x}right)}{12}+C$$
$$=dfrac{lnleft(mathrm{e}^{-24}left(mathrm{e}^{24}+1right)right)}{12}-dfrac{lnleft(2right)}{12}+1$$
which is rewritten as
$$dfrac{lnleft(mathrm{e}^{-24}left(mathrm{e}^{24}+1right)right)-lnleft(2right)+12}{12}$$
Is my work correct so far? Also, I thought I was done at the last step, but I noticed this integral can be approximated to $0.9422377349564838$. How would I go about doing this part?
If you want to approximate this numerically, we'll have to resort to series
$$frac{log cosh(12)}{12} = frac{logleft(frac{e^{12}+e^{-12}}{2e^{12}}right)+12}{12} approx frac{log(1-frac{1}{2})}{12}+1 approx 1 - frac{1}{24} - frac{1}{96}= frac{91}{96} sim 0.948$$
You won't need to consider the $e^{-12}$ until you want around $8$ figures of precision.
Answered by Ninad Munshi on January 7, 2022
Apply the limits to
$$dfrac{lnleft(mathrm{e}^{12x}+mathrm{e}^{-12x}right)}{12}bigg|_0^1= frac1{12}lnfrac{e^{12}+e^{-12}}{2}=frac1{12}ln cosh12 $$
Answered by Quanto on January 7, 2022
We have
$$frac{e^{ax}-e^{-ax}}{e^{ax}+e^{-ax}}=frac{sinh(ax)}{cosh(ax)}=tanh(ax)$$
Then the integral becomes
$$int_0^1 frac{e^{ax}-e^{-ax}}{e^{ax}+e^{-ax}}dx=int_0^1tanh(ax)dx=left. frac{ln(cosh(ax))}{a}right|_{x=0}^{x=1}=frac{ln(cosh(a))}{a}$$
With $a=12$ we get
$$int_0^1 frac{e^{12x}-e^{-12x}}{e^{12x}+e^{-12x}}dx=frac{ln(cosh(12))}{12}$$
Answered by QC_QAOA on January 7, 2022
$$int_0^1frac{e^{12x}-e^{-12x}}{e^{12x}+e^{-12x}} dx$$ $$=frac{1}{12}int_0^1frac{d(e^{12x}+e^{-12x})}{e^{12x}+e^{-12x}}$$ $$=frac{1}{12}[ln(e^{12x}+e^{-12x})]_0^1$$ $$=frac{1}{12}[ln(e^{12}+e^{-12})-ln(2)]$$
Answered by Harish Chandra Rajpoot on January 7, 2022
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