How to Simplify a Curve Integration Problem Using Trigonometric Identities

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SUMMARY

The discussion focuses on simplifying a curve integration problem using trigonometric identities. The original integral involves the expression {(x^3 + y) ds, with parameterizations x = 3t and y = t^3 over the interval 0 ≤ t ≤ 1. Participants suggest using trigonometric substitutions, specifically t^2 = 3cos(u) and t^2 = 3tan(u), to facilitate the integration process. The latter substitution is recommended as it aligns better with the identity sin^2(x) + cos^2(x) = 1, aiding in the simplification of the integral.

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  • Understanding of parametric equations and curve integration
  • Familiarity with trigonometric identities and substitutions
  • Knowledge of calculus, particularly integration techniques
  • Experience with differential calculus, specifically derivatives of parametric functions
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  • Study trigonometric substitutions in integral calculus
  • Learn about parametric equations and their applications in integration
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Students studying calculus, particularly those tackling curve integration problems, as well as educators looking for effective teaching methods in trigonometric applications in integration.

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Homework Statement


Well i got the formula down, I am having trouble with the integration though:
Q: {(x^3 + y) ds , x = 3t, y = t^3, 0<=t<=1

= {[ (27t^3 + t^3) sqrt(9t^2 + t^6)] dt


Homework Equations


f(x,y)ds = f(x(t), y(t)) sqrt[ (x'(t))^2 + (y'(t))^2 ]



The Attempt at a Solution


I simplified it down to: 28 {[ (t^4) sqrt(9 + t^4) ] dtAny help would be appreciated.

t^2 = 3cosu
= 28{ [(9cos^2u) sqrt(9 + 9cos^2u)] du
=756{ [(cos^2u) sqrt(1 + cos^2u)] du
no helpful identity..
dt^2 = -3sinu du
 
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WHY would you let t^2= 3 cos u? Since sin^2(x)+ cos^2(x)= 1, you have sin^2(x)= 1- cos^2(x) so "v= cos(x)" is useful for things like 1- cos^2(x), not 1+ cos^2(x).

Rather, dividing sin^2(x)+ cos^2(x)= 1 by cos^2(x) you have tan^2(x)+ 1= sec^2(x) so t^2= 3tan(u) would be much better.
 

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