# Evaluate $int_0^1frac{mathrm{e}^{12x}-mathrm{e}^{-12x}}{mathrm{e}^{12x}+mathrm{e}^{-12x}},mathrm{d}x$

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.

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