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anonymous

  • 4 years ago

need help with this calculus problems integrate (sqrt(9+0.0225t^2),0,22,t)dt

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  1. anonymous
    • 4 years ago
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    \[\int\limits_{0}^{22}\sqrt{9+0.0225t^2}dt\]

  2. anonymous
    • 4 years ago
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    there is a formula for that kind of form

  3. anonymous
    • 4 years ago
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    i dont know which method to use

  4. anonymous
    • 4 years ago
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    by substitution?

  5. anonymous
    • 4 years ago
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    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.

  6. dumbcow
    • 4 years ago
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    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

  7. anonymous
    • 4 years ago
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    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

  8. anonymous
    • 4 years ago
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    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))

  9. anonymous
    • 4 years ago
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    Substitute back for u = tan^(-1)(0.05 t)

  10. anonymous
    • 4 years ago
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    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

  11. anonymous
    • 4 years ago
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    Now with limits 0 and 20 it becomes ~~ 68.87

  12. anonymous
    • 4 years ago
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    OR http://www.wolframalpha.com/input/?i=integrate+%28sqrt%289%2B0.0225t%5E2%29dt

  13. anonymous
    • 4 years ago
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    thank you aron

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