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Math Help - Integration Help

  1. #1
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    Integration Help

    hi, im havin some problems solving this integral,

    if y= \int_0^\frac{1}{2\pi} \cos(x \sin \theta)  d\theta

    show that xy'' + y' +xy =0

    ive found the integral to be \frac{\sin(x \sin \theta)} {-x \cos\theta} is this correct, i found it using the substitution method where u = x \sin\theta

    if this integral is correct, how do i solve the rest of the equation?
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  2. #2
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    Quote Originally Posted by iLikeMaths View Post
    hi, im havin some problems solving this integral,

    if y= \int_0^\frac{1}{2\pi} \cos(x \sin \theta)  d\theta

    show that xy'' + y' +xy =0

    ive found the integral to be \frac{\sin(x \sin \theta)} {-x \cos\theta} is this correct, i found it using the substitution method where u = x \sin\theta

    if this integral is correct, how do i solve the rest of the equation?
    y= \int_0^\frac{1}{2\pi} \cos(x \sin \theta)  d\theta

    Use parts first.

    y= \int_0^\frac{1}{2\pi} \cos(x \sin \theta)  d\theta = \bigg(\theta   \cos(x \sin \theta)\bigg)\bigg|_0^\frac{1}{2\pi} - \int \theta (-\sin(x\sin\theta)x\cos(\theta))d\theta

     = \bigg(\theta \cos(x \sin \theta)\bigg)\bigg|_0^\frac{1}{2\pi} + \int \theta \sin(x\sin\theta)x\cos(\theta)d\theta

    Now use a substitution for the intergral.

     u = x\sin(\theta)

     \frac{du}{d\theta} = x\cos(\theta)

     du = x\cos(\theta)d\theta


     = \bigg(\theta \cos(x \sin \theta)\bigg)\bigg|_0^\frac{1}{2\pi} + \int \theta \sin(u)du

    And from  u = x\sin(\theta) , we can get  \theta = \asin\big\frac{u}{x}zbig)

    Integral is:

     = \bigg(\theta \cos(x \sin \theta)\bigg)\bigg|_0^\frac{1}{2\pi} + \int \arcsin \big(\frac{u}{x}\big) \sin(u)du
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  3. #3
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    when u integrated by parts, what did u use as ur "u" and "v" values because there seems to be a random \theta and i dont know where it came from, please explain, and also is that the complete integral or do i still have to integrate the part where there is \arcsinthanks for ur post
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