anonymous
  • anonymous
need help with this calculus problems integrate (sqrt(9+0.0225t^2),0,22,t)dt
Mathematics
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\int\limits_{0}^{22}\sqrt{9+0.0225t^2}dt\]
anonymous
  • anonymous
there is a formula for that kind of form
anonymous
  • anonymous
i dont know which method to use

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anonymous
  • anonymous
by substitution?
anonymous
  • anonymous
For the integrand, sqrt(0.0225 t^2+9) substitute t = 20. tan(u) and dt = 20. sec^2(u) du. Then sqrt(0.0225 t^2+9) = sqrt(9 tan^2(u)+9) = 3 sec(u) and u = tan^(-1)(0.05 t): = 60. integral sec^3(u) du Try using the reduction formula.
dumbcow
  • dumbcow
use the identity: 1 + tan^2 = sec^2 t = 3/sqrt(.0225) * tan u dt = 3/sqrt(.0225) * sec^2 u --> 9/sqrt(.0225) integral sec^3 du
anonymous
  • anonymous
The reduction formula: integral sec^m(u) du = (sin(u) sec^(m-1)(u))/(m-1) + (m-2)/(m-1) integral sec^(-2+m)(u) du, where m = 3
anonymous
  • anonymous
60. integral sec^3(u) du = 30. tan(u) sec(u)+30. integral sec(u) du = 30. tan(u) sec(u)+30. log(tan(u)+sec(u))
anonymous
  • anonymous
Substitute back for u = tan^(-1)(0.05 t)
anonymous
  • anonymous
That would be 1.5 sqrt(0.0025 t^2+1) t+30. log(sqrt(0.0025 t^2+1)+0.05 t) = 0.5 sqrt(0.0225 t^2+9.) t+30. sinh^(-1)(0.05 t)+constant
anonymous
  • anonymous
Now with limits 0 and 20 it becomes ~~ 68.87
anonymous
  • anonymous
OR http://www.wolframalpha.com/input/?i=integrate+%28sqrt%289%2B0.0225t%5E2%29dt
anonymous
  • anonymous
thank you aron

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