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

Question

2733 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-14T20:22:19

The arc element of the curve #y=f(x)# is given by the expression:

#dl = sqrt(dx^2+dy^2) = sqrt(dx^2+(f'(x)dx)^2) =dx sqrt(1+(f'(x))^2)#

as:

#f(x) = sqrt(cosx)#

#f'(x) = sinx/(2sqrt(cosx))#

integrating over the interval we have:

#l =int_(-pi/2)^(pi/2) sqrt (1+sin^2x/(2cosx))dx#