Finding arc length using integration

[subzero0137]

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

 

[dirk_mec1]

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

 

[subzero0137]

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 [itex]y=cosh^{-1}(x)[/itex].

 

[vela]

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



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

Looks fine to me.