# How do you find the arc length of the curve y=sqrt(cosx) over the interval [-pi/2, pi/2]?

Jan 14, 2017

The arc element of the curve $y = f \left(x\right)$ is given by the expression:

$\mathrm{dl} = \sqrt{{\mathrm{dx}}^{2} + {\mathrm{dy}}^{2}} = \sqrt{{\mathrm{dx}}^{2} + {\left(f ' \left(x\right) \mathrm{dx}\right)}^{2}} = \mathrm{dx} \sqrt{1 + {\left(f ' \left(x\right)\right)}^{2}}$

as:

$f \left(x\right) = \sqrt{\cos x}$

$f ' \left(x\right) = \sin \frac{x}{2 \sqrt{\cos x}}$

integrating over the interval we have:

$l = {\int}_{- \frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{1 + {\sin}^{2} \frac{x}{2 \cos x}} \mathrm{dx}$