Integration with a square root

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SUMMARY

The discussion centers on calculating the length of a parametrically defined curve given by the equations x = cos(t) and y = t + sin(t) for t in the interval [0, π]. The integral to evaluate is from 0 to π of the expression sqrt((dy/dt)² + (dx/dt)²), leading to the integral of sqrt(2 + 2cos(t)). Participants suggest using trigonometric identities, specifically the double-angle formula, to simplify the integral for easier computation. The recommended approach involves rewriting the expression using half-angle formulas and splitting the integral to handle absolute values.

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  • Understanding of parametric equations and curve length calculation
  • Knowledge of differentiation and integration techniques
  • Familiarity with trigonometric identities, particularly the double-angle and half-angle formulas
  • Basic calculus concepts, including definite integrals
NEXT STEPS
  • Learn how to apply trigonometric identities in integration, focusing on half-angle formulas
  • Study the method of integrating absolute value functions in definite integrals
  • Explore parametric equations and their applications in curve length calculations
  • Review techniques for simplifying integrals involving trigonometric functions
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Students studying calculus, particularly those focusing on parametric equations and integration techniques, as well as educators seeking to provide clarity on curve length calculations.

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Homework Statement


The problem is really about finding the length of a parametrically defined curve, this in itself seems easy enough, but it's the integration part that's hard.
The problem is as follows:
x = cost
y = t + sint
t = 0...pi


The Attempt at a Solution



So in order to find the length of this curve I need to find the integral from 0 to pi of:
sqrt((dy/dt)^2 + (dx/dt)^2)
dy/dt = 1 + cost
dx/dt = -sint

So I get the integral of:
sqrt((-sint)^2 + (1+cost)^2)) =
sqrt((sint)^2 + (cost)^2 + 2cost + 1)
(sint)^2 + (cost)^2 = 1, so I get the integral of:
sqrt(2+2cost).

And to be quite frank this has left me completely stumped. Googling around the internet I've read some stuff about trigonometric integration but there's nothing in the whole book about this, so I doubt that's how I'm supposed to solve it. I'm thinking I might be supposed to find a way to rewrite 2+2cost so that I can integrate but it has been to no avail.

A push in the right direction would definitely be appreciated.
 
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1+cos(t) has a nice expression in terms of a half angle formula.
 
you can use the double-angle formula: [tex]\cos^2 t=\frac{1+\cos 2t}{2}[/tex]

from there the integral is just a |cos (some multiple of t)| and you will need to split up the integral into 2 I think to get rid of the absolute value.
 

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