# Length of a curve

Find the length of the curve y=(e^x+e^-x)/2 from x=0 to x=ln4.
$$\frac{dy}{dx}= \frac{1}{2} \frac{d}{dx}e^x + \frac{1}{2} \frac{d}{dx}e^{-x}= \frac{1}{2}(e^x-e^{-x}) \\$$.
Therefore $$(\frac{dy}{dx})^2=\frac{1}{4}(e^x-e^{-x})\\$$ Therefore $$1+(\frac{dy}{dx})^2=\frac{1}{4}(e^{2x} -2e^xe^{-x}+e^{-2x}+4)^\frac{1}{2}\\$$.
Therefore $$\int_0^{ln4}\frac{1}{2}(e^{2x}-2e^xe^{-x}+e^{-2x}+4)^\frac{1}{2}dx$$. Do I solve this by substitution by letting $$u^2 = e^{2x}-2e^xe^{-x}+e{-2x}+4$$? Or what way should I approach it? Using the said substitution I get the following integral: $$\int u^2 \frac{1}{e^{2x}-e^{-2x}}du$$

cristo
Staff Emeritus
It will be a far simpler problem if you use the fact that $$\frac{e^x+e^{-x}}{2}=\cosh x$$. Have you encountered hyperbolic functions?

No I haven't encountered hyperbolic functions yet. They are two more chaphers away yet.Thanks for the help.

cristo
Staff Emeritus
No I haven't encountered hyperbolic functions yet. They are two more chaphers away yet.Thanks for the help.

Well, all you need to know for this question is that $$\frac{d}{dx}\cosh x=\sinh x , \hspace{1cm} \frac{d}{dx}\sinh x=\cosh x$$ and the identity $$\cosh^2x-\sinh^2x=1$$

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You said it would be a far simpler problem to use the fact that $$\frac{e^x+e^{-x}}{2}=coshx$$. Does that imply that it can still be done without the hyperbolic function method. If it does I would like to see the other longer way of solving this integral, please. Thanks.

Dick
Homework Helper
$$\int_0^{ln4}\frac{1}{2}(e^{2x}-2e^xe^{-x}+e^{-2x}+4)^\frac{1}{2}dx$$

Start here. e^x*e^(-x)=1, right?

Yes, I got as far as $$\int u^2\frac{1}{e^{x}-e^{-x}}du$$ but I cannot express $$e^x-e^{-x}$$ in terms of u.

Dick
Homework Helper
Then back up. It's much simpler than you are making it out to be. I'm trying to get you to notice that the expression inside of the (1/2) power in my last posting is a PERFECT SQUARE.

Yes, But what property of a perfect square am I to notice?

Dick
Homework Helper
That sqrt(a^2)=a. Simplify the integral before you start substituting.

$$\frac{1}{4}\int_0^{In4}((e^x + e^{-x}})^2)^\frac{1}{2}dx$$, is the integral to be solved.

Dick