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

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SUMMARY

The formula for finding the arclength of the curve defined by the integral of \( y = \int_{-\pi/2}^{x} \sqrt{\cos(t)} \, dt \) is derived using the arclength formula \( L = \int_{a}^{b} \sqrt{\left(\frac{dy}{dx}\right)^2 + 1} \, dx \). The derivative \( \frac{d}{dx} \int_{-\pi/2}^{x} \sqrt{\cos(t)} \, dt \) simplifies to \( \sqrt{\cos(x)} \) due to the Fundamental Theorem of Calculus, confirming that the constant limit does not affect the differentiation process. This establishes the necessary steps to compute the arclength for the specified interval.

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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?
 
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It is an application of the fondamental theorem of calculus.

\frac{d}{dx}\int_{x_0}^xf(t)dt=f(x)
 
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?
 
Last edited:
JJ6 said:
So does that mean that d/dx of the integral from -pi/2 to x of sqrt(cost)dt = sqrt(cosx)?

Yep.

JJ6 said:
Does the -pi/2 just disappear because it is a constant?
Disapear from where?
 

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