Answer
$\approx{3.820}$
Work Step by Step
The formula for the arc length is
$s=\int_{a}^b \sqrt{1 + (y')^2} dy$
Differentiate $y = \cos{x}$ with respect to x.
$y' =-\sin{x}$
Substitute the value of y' in $1 + (y')^2$
$1+(y')^2=1+{\sin{^2}{x}}$
Substitute the value of $1 + (y')^2$ in the formula for the arc length $s=\int_{a}^b \sqrt{1 + (y')^2} dy$
Solve for the distance s
$s==\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\sqrt{1+\sin{^2}{x}} dx$]
$\approx{3.820}$
