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Homework Help: Don't know to handle the (ωt) and (t) that appear in the integral

  1. May 4, 2013 #1
    1. The problem statement, all variables and given/known data
    I have the integral:
    [tex]\int_{0}^{4} \sin(\omega t)\cdot t^2[/tex]

    2. Relevant equations
    I know from integrals tables that:
    [tex]\int x^2 \sin x=2x\sin x-(x^2-2)\cos x[/tex]
    But i don't know to handle the (ωt) and (t) that appear in the integral

    3. The attempt at a solution
  2. jcsd
  3. May 4, 2013 #2
    Very simple: $$

    \int_{0}^{4} \sin(\omega t)\cdot t^2 d \omega = t^2 \int_0^4 \sin (\omega t) d \omega

    = t^2 \left[ \frac {-\cos (\omega t)} {t} \right]_0^4 = t (1 - \cos 4t)
  4. May 4, 2013 #3
    Sorry, i didn't mention it's not dω, but dt, the variant is the time, the ω is a constant
  5. May 4, 2013 #4
    Hint: Write [itex] t^2= \frac{1}{\omega^2}(\omega t)^2 [/itex].
  6. May 4, 2013 #5
    Do you see why it is important to append dx to the integrand? Without this, no one can be sure what variable is being integrated.
  7. May 4, 2013 #6
    So, i use [itex] t^2= \frac{1}{\omega^2}(\omega t)^2 [/itex], and then i define a new variable x=ωt and then dx=ωdt, right?
  8. May 4, 2013 #7
    That should work.
  9. May 4, 2013 #8


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    Simplest is to use the substitution [itex]y= \omega t[/itex] so that [itex]dy= \omega dt[/itex], [itex]dt= dy/omega[/itex] and [itex]t= y/\omega[/itex]
    Last edited by a moderator: May 4, 2013
  10. May 5, 2013 #9

    The integral is:
    [tex]\int_{0}^{4} \sin(\omega t)\cdot t^2[/tex]
    I used:
    [tex]t^2= \frac{1}{\omega^2}(\omega t)^2 [/tex]
    Then x=ωt and then dx=ωdt.
    for t=o --> x=0
    for t=4 --> x=ωx4[sec]=0.00029
    From integrals tables:
    [tex]\int x^2 \sin x=2x\sin x-(x^2-2)\cos x[/tex]
    [tex]\frac{1}{\omega^3}\int_{0}^{0.00029}\sin x-(x^2-2)\cos x=211093[/tex]
    Can anyone check? the result isn't logical, physically
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