Question

need help with this calculus problems

integrate (sqrt(9+0.0225t^2),0,22,t)dt

Answer 1

\[\int\limits_{0}^{22}\sqrt{9+0.0225t^2}dt\]

Answer 2

there is a formula for that kind of form

Answer 3

i dont know which method to use

Answer 4

by substitution?

Answer 5

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.

Answer 6

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

Answer 7

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

Answer 8

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))

Answer 9

Substitute back for u = tan^(-1)(0.05 t)

Answer 10

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

Answer 11

Now with limits 0 and 20 it becomes ~~ 68.87

Answer 12

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

Answer 13

thank you aron