# Finding arc length using integration

Find the length of the positive arc of the curve $y=cosh^{-1}(x)$ (for which y≥0) between $x=1$ and $x=\sqrt{5}$.

My attempt: $x=cosh(y) → \frac{dx}{dy} = sinh(y) → (\frac{dx}{dy})^{2}=sinh^{2}(y)$, so $ds=dy\sqrt{1+sinh^{2}(y)}$, therefore the arc length is $S=\int_{y=0}^{y=cosh^{-1}(\sqrt{5})} cosh(y) dy= 2$. Is this right? Even if it is, is there another method of doing it (e.g. parametric equations)?

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No you have to integrate between 1 and sqrt(5).

No you have to integrate between 1 and sqrt(5).

But those are the x limits. If I want to integrate with respect to dy, I need y limits. I could've used dy/dx instead of dx/dy, but I don't know how to differentiate $y=cosh^{-1}(x)$.

vela
Staff Emeritus
Find the length of the positive arc of the curve $y=cosh^{-1}(x)$ (for which y≥0) between $x=1$ and $x=\sqrt{5}$.
My attempt: $x=cosh(y) → \frac{dx}{dy} = sinh(y) → (\frac{dx}{dy})^{2}=sinh^{2}(y)$, so $ds=dy\sqrt{1+sinh^{2}(y)}$, therefore the arc length is $S=\int_{y=0}^{y=cosh^{-1}(\sqrt{5})} cosh(y) dy= 2$. Is this right? Even if it is, is there another method of doing it (e.g. parametric equations)?