What is the formula for finding arclength when given an integral?

[JJ6]

Homework Statement

Find the arclength of the curve given by y= integral from -pi/2 to x of sqrt(cost)dt. X is restricted between -pi/2 and pi/2.

Homework Equations

L = Integral from a to b of sqrt((dy/dx)^2 + 1)dx
L = Integral from a to b of sqrt((dy/dt)^2 + (dx/dt)^2)dt

The Attempt at a Solution

I'm not even sure how to start this problem since integral of sqrt(cost) has no simple function representing it. Can somebody please give me some direction?

 

[quasar987]

It is an application of the fondamental theorem of calculus.

$$\frac{d}{dx}\int_{x_0}^xf(t)dt=f(x)$$

 

[JJ6]

So does that mean that d/dx of the integral from -pi/2 to x of sqrt(cost)dt = sqrt(cosx)?

Does the -pi/2 just disappear because it is a constant?

 

[quasar987]

So does that mean that d/dx of the integral from -pi/2 to x of sqrt(cost)dt = sqrt(cosx)?

Yep.

Does the -pi/2 just disappear because it is a constant?

Disapear from where?