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Math Help - help with integration

  1. #1
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    help with integration

    \int_{0}^{2} t^2sin{t^3} + 2.5dt
    So I started with integration by parts.
    \int_{0}^{2} t^2sint^3dt + \int_{0}^{2} 2.5dt
     u = t^2
     du = 2tdt
     dv = sint^3dt
     v = \frac{-1}{3t^2}cos(t^3)
    but then I realized this was gonna go on for a long time. Is there an easier way to take this integral?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by steveo0 View Post
    \int_{0}^{2} t^2sin{t^3} + 2.5dt
    So I started with integration by parts.
    \int_{0}^{2} t^2sint^3dt + \int_{0}^{2} 2.5dt
     u = t^2
     du = 2tdt
     dv = sint^3dt
     v = \frac{-1}{3t^2}cos(t^3)
    but then I realized this was gonna go on for a long time. Is there an easier way to take this integral?
    If you mean \int_0^2t^2\sin\left(t^3\right)+2.5\,dt, then we see that we have \int_0^2t^2\sin\left(t^3\right)\,dt+2.5\int_0^2\,d  t

    To solve the first integral, make the substitution u=t^3
    The second one is very straightforward.

    Can you continue from here?
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  3. #3
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    oh! yeah. thanks
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  4. #4
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    Hello, steveo0!

    By-parts is not necessary . . .

    We are concerned with: . \int_{0}^{2} t^2\sin(t^3)\,dt


    Let: . u \,=\,t^3\quad\Rightarrow\quad du \,=\,3t^2\,dt \quad\Rightarrow\quad t^2\,dt \,=\,\tfrac{1}{3}du


    We have: . \int \underbrace{\sin(t^3)}_{\downarrow}\,\underbrace{(  t^2\,dt)}_{\downarrow}
    Substitute: . \int \sin u\,\left(\tfrac{1}{3}du \right) \;=\;\tfrac{1}{3}\int \sin u\,du . . . Got it?

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