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Math Help - how to integrate sin(x/y) dx

  1. #1
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    how to integrate sin(x/y) dx

    can anyone assist with evaluating the integral if sin (x/y) with respect to x & y, i have tried using integration by parts but cant seem to get an answer.

    i know by using the online calculators the answer is -ycos(x/y)
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    Quote Originally Posted by john1985 View Post
    can anyone assist with evaluating the integral if sin (x/y) with respect to x & y, i have tried using integration by parts but cant seem to get an answer.

    i know by using the online calculators the answer is -ycos(x/y)
    When you integrate with respect to x, treat y as a constant.
    When you integrate with respect to y, treat x as a constant
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    suggestion

    Quote Originally Posted by john1985 View Post
    can anyone assist with evaluating the integral if sin (x/y) with respect to x & y, i have tried using integration by parts but cant seem to get an answer.

    i know by using the online calculators the answer is -ycos(x/y)
    t=x/y.
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    would some be able to post some of the initial steps i get you have to use the substitution method but am having difficultiy getting the answer out

    original expression integrate with respect to x
    = sin(y/x)

    step 2 using substitution = sin (t) dt

    step 3 -cos (t) t ???
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  5. #5
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    Is y a constant? If so: let x = yt, so dx = y dt. Then \int \sin (\frac xy)\;\mathrm{d}x = \int \sin (t) \cdot y\;\mathrm{d}t = y \int \sin (t) \;\mathrm{d}t = -y \cos t + c = -y\cos(\frac xy) + c.
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    Quote Originally Posted by john1985 View Post
    can anyone assist with evaluating the integral if sin (x/y) with respect to x & y, i have tried using integration by parts but cant seem to get an answer.

    i know by using the online calculators the answer is -ycos(x/y)
    You might not think that the region of integration is important, but it is. Post the whole question please if you hope to get effective help.
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    \int sin \bigg( \frac{x}{y} \bigg)dx=y\int sin \bigg(\frac{x}{y} \bigg)\bigg(\frac{1}{y}dx\bigg)=-ycos\bigg(\frac{x}{y}\bigg)+C

    Try doing dy now.
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  8. #8
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    Quote Originally Posted by dwsmith View Post
    \int sin \bigg( \frac{x}{y} \bigg)dx=y\int sin \bigg(\frac{x}{y} \bigg)\bigg(\frac{1}{y}dx\bigg)=-ycos\bigg(\frac{x}{y}\bigg)+C

    Try doing dy now.
    I doubt this is what the question has asked the OP to do. See the boldface in the quote below:

    Quote Originally Posted by john1985 View Post
    can anyone assist with evaluating the integral if sin (x/y) with respect to x & y, i have tried using integration by parts but cant seem to get an answer.

    i know by using the online calculators the answer is -ycos(x/y)
    Until what I've boldfaced is explained, the real question cannot be reliably answered.
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  9. #9
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    Quote Originally Posted by mr fantastic View Post
    I doubt this is what the question has asked the OP to do. See the boldface in the quote below:



    Until what I've boldfaced is explained, the real question cannot be reliably answered.
    I am under the impression that he wants to integrate with dx and then a separate integral of dy.
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  10. #10
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    Quote Originally Posted by dwsmith View Post
    I am under the impression that he wants to integrate with dx and then a separate integral of dy.
    Doubtful, since integrating again w.r.t. y will require non-elementary functions...
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  11. #11
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    Quote Originally Posted by dwsmith View Post
    I am under the impression that he wants to integrate with dx and then a separate integral of dy.
    I'll bet dollars to doughnuts that the original question gives a region that has to be integrated over. But we will never know unless the OP replies. Further posts are useless until then.
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    evaluate the double integral sin(x/y)dA where R is the region bounded by the y-axis , y=pi and x=y^2
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  13. #13
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    Quote Originally Posted by sr917 View Post
    evaluate the double integral sin(x/y)dA where R is the region bounded by the y-axis , y=pi and x=y^2
    Draw the region of integration. It is then clear that the required integral is

    \int_{y = 0}^{y = \pi} \int_{x = 0}^{x = y^2} \sin \left(\frac{x}{y}\right) \, dx \, dy

     = \int_0^\pi \left[-y \cos \left( \frac{x}{y}\right) \right]_0^{y^2} \, dy

     = - \int_0^\pi y \cos (y) \, dy = ....

    (If only the complete question had been asked in the first place a lot of time and energy would not have been wasted).
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