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Thread: Evaluating Irrational Integral

  1. #1
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    Post Evaluating Irrational Integral

    I need help to integrate this particular integral step-by-step. I've been wrestling around with substitution method without any satisfying result. I don't know I'm doing wrong with it.

    \int_{-a}^a \frac{1}{\left(x^2+y^2\right)^{3/2}} \, dy

    This is my attempt:
    Let u = \left(x^2+y^2\right)
    du = 2y\text{dy}

    Hence:
    \int \frac{1}{\left(x^2+y^2\right)^{3/2}} \, dy<br />
\int \frac{u^{-3/2}}{2} \, du<br />
\frac{1}{2} \int u^{-3/2} \, du<br />
\frac{\frac{1}{\sqrt{u}}}{\frac{2 (-1)}{2}}-\frac{1}{\sqrt{u}}-\frac{1}{\sqrt{x^2+y^2}}-\frac{1}{\sqrt{x^2+y^2}}
    -\frac{1}{\sqrt{x^2+y^2}}

    Thank You
    Last edited by Lites; Nov 23rd 2011 at 04:59 AM. Reason: Solved in Physics forum.
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  2. #2
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    Re: Evaluating Irrational Integral

    Quote Originally Posted by Lites View Post
    I need help to integrate this particular integral step-by-step. I've been wrestling around with substitution method without any satisfying result. I don't know I'm doing wrong with it.

    \int_{-a}^a \frac{1}{\left(x^2+y^2\right)^{3/2}} \, dy

    This is my attempt:
    Let u = \left(x^2+y^2\right)
    du = 2y\text{dy}

    Hence:
    \int \frac{1}{\left(x^2+y^2\right)^{3/2}} \, dy<br />
\int \frac{u^{-3/2}}{2} \, du<br />
\frac{1}{2} \int u^{-3/2} \, du<br />
\frac{\frac{1}{\sqrt{u}}}{\frac{2 (-1)}{2}}-\frac{1}{\sqrt{u}}-\frac{1}{\sqrt{x^2+y^2}}-\frac{1}{\sqrt{x^2+y^2}}
    -\frac{1}{\sqrt{x^2+y^2}}

    Thank You
    note that since du = 2y \, dy , dy = \frac{du}{2y} ... you left out the y in the denominator in your substitution.

    try a trig substitution, maybe?

    y = x\tan{t}
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