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Int sin[sqrt[t]]

  1. Sep 16, 2005 #1
    I'm drawing a blank as to how to approach this one. I've been looking over all trig identities and substitutions I could possibly make - but to no avail :cry:.

    \int_{1}^{x^2} \sin{(\sqrt{t})} \ dt
  2. jcsd
  3. Sep 16, 2005 #2
    Try making a u-substitution for the square root of t.
  4. Sep 17, 2005 #3
    indefinite integral...

    The closest identity that I could determine is:

    [tex]\int u \sin u \; du = \sin u - u \cos u + C[/tex]

    [tex]\int \sin \sqrt{t} \; dt = 2 \left( \sin \sqrt{t} - \sqrt{t} \cos \sqrt{t} \right) + C[/tex]
  5. Sep 18, 2005 #4


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    Good try but it looks to me like that gives you
    [tex]\int u^{\frac{1}{2}}sin u du[/tex] which doesn't look any more hopeful.

    I'd be willing to bet that this doesn't have an elementary anti-derivative.
  6. Sep 19, 2005 #5


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    Don't bet too much, Halls.

    [tex] \sqrt{t}=u [/tex]

    implies [itex] t=u^{2} \ \mbox{and} \ dt= 2 u du [/itex]

    and the antiderivative becomes

    [tex] \int 2 u \sin u \ du [/tex] which can be easily tackled with the part integration method.

  7. Sep 20, 2005 #6


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    One of these days, I really have to learn algebra!
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