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Math Help - Easy trig integral =/

  1. #1
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    Easy trig integral =/

    I haven't done any sort of trig integrals in the past and i have pretty much forgotten all my identities since i haven't looked at them for over 3 years so please bear with me.

    The question is:

    upper limit = 1, lower limit = 0.
    Integrate cos (x^2) with those bounds.

    First off I had to approximate the integral using the midpoint/trapezoidal rules. Those were fairly straightforward. Now i want to check my answer but I'm kinda lost.

    If i took the integral of cos (x^2), i would get sin(x^2) no?
    Subbing in the limits of integration i would get sin(1) - sin(0) = sin (1)
    but using the approximation rules above I got the integral as being approximately equal to one, which does not equal to sin (1).

    Help?
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  2. #2
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    Quote Originally Posted by Kuma View Post
    I haven't done any sort of trig integrals in the past and i have pretty much forgotten all my identities since i haven't looked at them for over 3 years so please bear with me.

    The question is:

    upper limit = 1, lower limit = 0.
    Integrate cos (x^2) with those bounds.

    First off I had to approximate the integral using the midpoint/trapezoidal rules. Those were fairly straightforward. Now i want to check my answer but I'm kinda lost.

    If i took the integral of cos (x^2), i would get sin(x^2) no? Mr F says: No. Differentiate it and you will see why.
    Subbing in the limits of integration i would get sin(1) - sin(0) = sin (1)
    but using the approximation rules above I got the integral as being approximately equal to one, which does not equal to sin (1).

    Help?
    The integral of cos(x^2) cannot be found in closed form using elementary functions.
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  3. #3
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    Thanks. But how would i differentiate that then? Does chain rule apply or something. I've never worked with trig functions before so. I do know that d/dx cos x = sin x, how would that apply to cos x^2?
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  4. #4
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    Yeah, you use the chain rule. First, find the derivative of sin, then find the derivative of the value within sin.



    Easy trig integral =/-capture4.jpg
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  5. #5
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    Ah i see. So you can differentiate cos x^2 to get 2x sin x^2, but the integral of that would pretty much be impossible to find. Makes sense.
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  6. #6
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    I find that \int_{0}^{1}\cos{x^2}\;{dx} = \sum\limits_{k=0}^{\infty}\frac{(-1)^k}{(2k)!(4k+1)}.
    (Not that it's significant or matters, anyway).
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