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Math Help - Solving this double integral.

  1. #1
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    Solving this double integral.

    Hi.

    I want to solve this.

    Solving this double integral.-j4mc6.png

    I'm not sure what to do. I used integration by parts and I got stuck. I tried changing the order as well and i think it made it worse.

    set:

    u = 1/y
    du = -1/y^2
    dv = cos y
    v = sin y

    so i get

    sin y/y + int (cos y)/y^2 from integration by parts.

    If i change the order, I get the bounds as 0 to 1 for the outer integral and y to 1 for the inner integral and switching the dx and dy. That gives me

    (cos y)/y - cos y for the inner integral.

    So I'm not sure what to do now...
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  2. #2
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    Re: Solving this double integral.

    Quote Originally Posted by Kuma View Post
    Hi.

    I want to solve this.

    Click image for larger version. 

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    I'm not sure what to do. I used integration by parts and I got stuck. I tried changing the order as well and i think it made it worse.

    set:

    u = 1/y
    du = -1/y^2
    dv = cos y
    v = sin y

    so i get

    sin y/y + int (cos y)/y^2 from integration by parts.

    If i change the order, I get the bounds as 0 to 1 for the outer integral and y to 1 for the inner integral and switching the dx and dy. That gives me

    (cos y)/y - cos y for the inner integral.

    So I'm not sure what to do now...
    The first thing you need to do is change the order of integration.

    From your bounds, you can see \displaystyle 0 \leq x \leq 1 and \displaystyle x \leq y \leq 1.

    It should be clear therefore that \displaystyle 0 \leq x \leq y and \displaystyle 0 \leq y \leq 1.

    So your double integral becomes

    \displaystyle \begin{align*} \int_0^1{\int_x^1{\frac{\cos{y}}{y}\,dy}\,dx} &= \int_0^1{\int_0^y{\frac{\cos{y}}{y}\,dx}\,dy} \\ &= \int_0^1{\left[\frac{x\cos{y}}{y}\right]_0^y\,dy} \\ &= \int_0^1{\left(\frac{y\cos{y}}{y} - \frac{0\cos{y}}{y}\right)\,dy} \\ &= \int_0^1{\cos{y}\,dy} \\ &= \left[\sin{y}\right]_0^1 \\ &= \sin{1} - \sin{0} \\ &= \sin{1} \end{align*}
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