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Thread: Integration Help

  1. #1
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    Integration Help

    hi, im havin some problems solving this integral,

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

    show that $\displaystyle xy'' + y' +xy =0$

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

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

    show that $\displaystyle xy'' + y' +xy =0$

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

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

    Use parts first.

    $\displaystyle 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 $

    $\displaystyle = \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.

    $\displaystyle u = x\sin(\theta) $

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

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


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

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

    Integral is:

    $\displaystyle = \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 $\displaystyle \theta$ 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 $\displaystyle \arcsin$thanks for ur post
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