Question 972247: To help find the velocity of particles requires the evaluation of the indefinite
integral of the acceleration function, a(t), i.e. Z replaced integral sign
v =Z a(t) dt
Evaluate the following indefinite integrals
Z 3lnt/(t^3) dt
and
Z (t^2 + t) cos3tdt
Amy.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! let Z be the integral sign, then integration by parts is defined
Z udv = uv - Z vdu
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a) Z 3lnt/(t^3) dt
let u = 3ln(t) and dv = t^-3, then du = 3/t, v = Z t^-3 = -1/(2t^2), therefore
Z 3lnt/(t^3) dt = -3ln(t)/(2t^2) - Z -3/(2t^3) + C
Z 3lnt/(t^3) dt = -3ln(t)/(2t^2) + Z 3/(2t^3) + C
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b) Z (t^2 + t) cos3tdt
let u = (t^2 + t) and dv = cos3t, then du = 2t+1, v = Z cos3t = (1/3)sin(3t), therefore
Z (t^2 + t) cos3tdt = (t^2 + t)(1/3)sin(3t) - Z (1/3)sin(3t)(2t+1) + C
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