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  1. #1
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    integral question

    How do I start this integral?
    integral from [-2 to 1]
    sqrt((x^2)-1) / x dx
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  2. #2
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    Re: integral question

    Quote Originally Posted by Sneaky View Post
    How do I start this integral?
    integral from [-2 to 1]
    sqrt((x^2)-1) / x dx
    \displaystyle \int_{-2}^1{\frac{\sqrt{x^2 - 1}}{x}\,dx} = \frac{1}{2}\int_{-2}^1{\frac{2x\sqrt{x^2 - 1}}{x^2}\,dx}

    Now make the substitution \displaystyle u = x^2 - 1 \implies du = 2x\,dx, and making note that \displaystyle u(-2) = 3, u(1) = 0, the integral becomes

    \displaystyle \begin{align*} \frac{1}{2}\int_{-2}^1{\frac{2x\,\sqrt{x^2 - 1}}{x^2}\,dx} &= \frac{1}{2}\int_3^0{\frac{\sqrt{u}}{u + 1}\,du} \\ &= -\frac{1}{2}\int_0^3{\frac{\sqrt{u}}{u+1}\,du} \\ &= -\int_0^3{\frac{u}{2\sqrt{u}\,(u + 1)}\,du} \end{align*}

    Now make the substitution \displaystyle v = \sqrt{u} \implies dv = \frac{1}{2\sqrt{u}}\,du and noting that \displaystyle v(0) = 0 and \displaystyle v(3) = \sqrt{3}, the integral becomes

    \displaystyle \begin{align*} -\int_0^3{\frac{u}{2\sqrt{u}\,(u + 1)}\,du} &= -\int_0^{\sqrt{3}}{\frac{v^2}{v^2 + 1}\,dv} \\ &= -\int_0^{\sqrt{3}}{1 - \frac{1}{v^2 + 1}\,dv} \\ &= -\left[v - \arctan{v}\right]_0^{\sqrt{3}} \\ &= -\left[\left(\sqrt{3} - \arctan{\sqrt{3}}\right) - \left(0 - \arctan{0}\right)\right] \\ &= -\left(\sqrt{3} - \frac{\pi}{3} - 0 + 0\right) \\ &= \frac{\pi}{3} - \sqrt{3} \end{align*}
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  3. #3
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    Re: integral question

    Quote Originally Posted by Sneaky View Post
    How do I start this integral?
    integral from [-2 to 1]
    sqrt((x^2)-1) / x dx
    please recheck the given limits of integration. the integrand, \frac{\sqrt{x^2-1}}{x}, is undefined on the interval -1 < x < 1
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    Re: integral question

    oh i mis-read it, its -2 to -1. But I think I can fix it myself.
    Also I have a question about how to change the integral bounds when you do a change of variable. What if you change u=cos(x), and the integral bound was -inf to inf, then what would be the new integral bounds?
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    Re: integral question

    Quote Originally Posted by Sneaky View Post
    oh i mis-read it, its -2 to -1. But I think I can fix it myself.
    Also I have a question about how to change the integral bounds when you do a change of variable. What if you change u=cos(x), and the integral bound was -inf to inf, then what would be the new integral bounds?
    I expect you would not ever have such an integral...
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  6. #6
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    Re: integral question

    For the integral ln(x) / x dx, i let u=ln(x), then got
    [e to infinity]

    integral u du
    [1 to infinity]

    then i got (after integrating and subbing in the bounds)
    ln^2(infinity)/2 - 0

    Is this right so far?
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  7. #7
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    Re: integral question

    If you change the bounds, you don't convert the function back to a function of x, leave it in terms of u and substitute the u values.

    But you are correct that the integral does not converge.
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  8. #8
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    Re: integral question

    If i do that I end up with (u^2) / 2 |[1 to inf]
    then it ends up being infinity - 1/2
    It doesn't look correct.
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    Re: integral question

    Does this mean the answer is infinity?
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  10. #10
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    Re: integral question

    Quote Originally Posted by Sneaky View Post
    Does this mean the answer is infinity?
    It means that the integral does not converge... Infinity is not a number...
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