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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{22}\sqrt{9+0.0225t^2}dt\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there is a formula for that kind of form

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i dont know which method to use

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0use 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
 4 years ago
Best ResponseYou've already chosen the best response.0The reduction formula: integral sec^m(u) du = (sin(u) sec^(m1)(u))/(m1) + (m2)/(m1) integral sec^(2+m)(u) du, where m = 3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.060. 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
 4 years ago
Best ResponseYou've already chosen the best response.0Substitute back for u = tan^(1)(0.05 t)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That 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
 4 years ago
Best ResponseYou've already chosen the best response.0Now with limits 0 and 20 it becomes ~~ 68.87

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0OR http://www.wolframalpha.com/input/?i=integrate+%28sqrt%289%2B0.0225t%5E2%29dt
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