Length of Curve: Evaluating the Integral

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SUMMARY

The discussion focuses on evaluating the integral to find the length of the curve defined by the parametric equations x=cos(2t) and y=sin(3t) over the interval 0≤t≤2π. The user correctly identifies that the length can be calculated using the integral of the magnitude of the velocity vector. The derivative calculations yield x'=-2sin(2t) and y'=3cos(3t), leading to the integral c=∫√(4sin²(2t)+9cos²(3t)) dt. The user seeks assistance in evaluating this integral, particularly after simplifying it to the form 4 + 5cos²(3t).

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Homework Statement


Find the length of the curve.

x=cos(2t)
y=sin(3t)
0≤t≤2∏

I know the length is just the integral from 0 to 2∏of the magnitude of the velocity.


The Attempt at a Solution



x'=-2sin(2t) x'^2=4sin^2(2t)
y'=3cos(3t) y'^2=9cos^2(3t)

c=∫√4sin^2(2t)+9cos^2(3t)

Im having trouble evaluating the integral. Could some please point me in the right direction?

Thanks
 
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[itex]4sin^2(3t)+ 9cos^2(3t)= 4sin^2(3t)+ 4cos^2(3t)+ 5cos^2(3t)= 4+ 5cos^2(3t)[/itex]. Let [itex]u= 4+ 5cos^2(3t)[/itex].
 

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