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Math Help - Problems trying to solve an integral

  1. #1
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    Problems trying to solve an integral

    Hi everyone,
    I need to solve this integral, but I get stuck in some point and I can't go on. I explain what I have been trying:

    \displaystyle\int\sqrt[ ]{\frac{u^2-x^2}{u^2-a^2}}\frac{a}{u}du

    Changing the variable as

    t^2=\displaystyle\frac{u^2-x^2}{u^2-a^2}

    So "u" would be:

    t^2(u^2-a^2)=u^2-x^2 Desarrollando esta expresión,llegamos a que:

    u^2(t^2-1)=t^2a^2-x^2 Por lo que u=\sqrt[ ]{\displaystyle\frac{t^2a^2-x^2}{t^2-1}}

    The derivative of u with respect to: \frac{du}{dt}

    \displaystyle\frac{du}{dt}=\displaystyle\frac{1}{2  }\displaystyle\frac{1}{\sqrt[ ]{\displaystyle\frac{t^2a^2-x^2}{t^2-1}}}\displaystyle\frac{(t^2-1)(2ta^2)-(t^2a^2-x^2)(2t)}{(t^2-1)^2}

    \displaystyle\frac{du}{dt}=\displaystyle\frac{1}{2  }\displaystyle\frac{1}{\sqrt[ ]{\displaystyle\frac{t^2a^2-x^2}{t^2-1}}}\displaystyle\frac{2t(x^2-a^2)}{(t^2-1)^2}

    Replacing these terms in the original integral we have:

    \displaystyle\int\sqrt{\frac{u^2-x^2}{u^2-a^2}}\frac{a}{u}du=\displaystyle\int\sqrt{t^2}\dis  \frac{a}{\sqrt{\frac{t^2a^2-x^2}{t^2-1}}}\frac{1}{\sqrt{\frac{t^2a^2-x^2}{t^2-1}}}\frac{2t(x^2-a^2)}{(t^2-1)^2}dt

    \displaystyle\int\frac{t^2(x^2-a^2)a}{\left(\dfrac{t^2a^2-x^2}{t^2-1}\right)(t^2-1)^2}dt

    When I get to this point, I don't know how to go on. The result of the integral would be like this:

    \displaystyle\int\frac{t^2(x^2-a^2)a}{\left(\dfrac{t^2a^2-x^2}{t^2-1}\right)(t^2-1)^2}dt=\frac{a}{2}\ln\left(\frac{t+1}{t-1}\right)-\frac{x}{2}\ln\left(\frac{t+(x/a)}{t-(x/a)}\right)

    I tried operating in other way, like this:

    \displaystyle\int\sqrt[ ]{\frac{u^2-x^2}{u^2-a^2}}\frac{a}{u}du

    Doing the change of variable t^2=\displaystyle\frac{u^2-x^2}{u^2-a^2} we have this:

    2tdt=\displaystyle\frac{2u(u^2-a^2)-2u(u^2-x^2)}{(u^2-a^2)^2}du=u\displaystyle\frac{(x^2-a^2)}{(u^2-a^2)^2}du

    \displaystyle\frac{u}{(u^2-a^2)^2}du=\displaystyle\frac{t}{(x^2-a^2)}dt

    But again, I don't know how to continue.

    Could you tell me what could I do to go on with the integral?
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  2. #2
    Junior Member bondesan's Avatar
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    Re: Problems trying to solve an integral

    If you factor the denominator u^2-a^2 into u^2\left(1-\left(\frac{a}{u}\right)^2\right), you can play with some change of variables, since you have a a/u outside the root. But I don't know if this is going to work. Just my 2 cents.
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  3. #3
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    Re: Problems trying to solve an integral

    Quote Originally Posted by nekovolta View Post
    Hi everyone,
    I need to solve this integral, but I get stuck in some point and I can't go on. I explain what I have been trying:

    \displaystyle\int\sqrt[ ]{\frac{u^2-x^2}{u^2-a^2}}\frac{a}{u}du

    Changing the variable as

    t^2=\displaystyle\frac{u^2-x^2}{u^2-a^2}

    So "u" would be:

    t^2(u^2-a^2)=u^2-x^2 Desarrollando esta expresión,llegamos a que:

    u^2(t^2-1)=t^2a^2-x^2 Por lo que u=\sqrt[ ]{\displaystyle\frac{t^2a^2-x^2}{t^2-1}}

    The derivative of u with respect to: \frac{du}{dt}

    \displaystyle\frac{du}{dt}=\displaystyle\frac{1}{2  }\displaystyle\frac{1}{\sqrt[ ]{\displaystyle\frac{t^2a^2-x^2}{t^2-1}}}\displaystyle\frac{(t^2-1)(2ta^2)-(t^2a^2-x^2)(2t)}{(t^2-1)^2}

    \displaystyle\frac{du}{dt}=\displaystyle\frac{1}{2  }\displaystyle\frac{1}{\sqrt[ ]{\displaystyle\frac{t^2a^2-x^2}{t^2-1}}}\displaystyle\frac{2t(x^2-a^2)}{(t^2-1)^2}

    Replacing these terms in the original integral we have:

    \displaystyle\int\sqrt{\frac{u^2-x^2}{u^2-a^2}}\frac{a}{u}du=\displaystyle\int\sqrt{t^2}\dis  \frac{a}{\sqrt{\frac{t^2a^2-x^2}{t^2-1}}}\frac{1}{\sqrt{\frac{t^2a^2-x^2}{t^2-1}}}\frac{2t(x^2-a^2)}{(t^2-1)^2}dt

    \displaystyle\int\frac{t^2(x^2-a^2)a}{\left(\dfrac{t^2a^2-x^2}{t^2-1}\right)(t^2-1)^2}dt

    When I get to this point, I don't know how to go on. The result of the integral would be like this:

    \displaystyle\int\frac{t^2(x^2-a^2)a}{\left(\dfrac{t^2a^2-x^2}{t^2-1}\right)(t^2-1)^2}dt=\frac{a}{2}\ln\left(\frac{t+1}{t-1}\right)-\frac{x}{2}\ln\left(\frac{t+(x/a)}{t-(x/a)}\right)

    I tried operating in other way, like this:

    \displaystyle\int\sqrt[ ]{\frac{u^2-x^2}{u^2-a^2}}\frac{a}{u}du

    Doing the change of variable t^2=\displaystyle\frac{u^2-x^2}{u^2-a^2} we have this:

    2tdt=\displaystyle\frac{2u(u^2-a^2)-2u(u^2-x^2)}{(u^2-a^2)^2}du=u\displaystyle\frac{(x^2-a^2)}{(u^2-a^2)^2}du

    \displaystyle\frac{u}{(u^2-a^2)^2}du=\displaystyle\frac{t}{(x^2-a^2)}dt

    But again, I don't know how to continue.

    Could you tell me what could I do to go on with the integral?
    Is this integral supposed to be w.r.t. u or w.r.t. x?
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  4. #4
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    Re: Problems trying to solve an integral

    Hi, the integral is with respect to "u".
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  5. #5
    Ted
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    Re: Problems trying to solve an integral

    Can you tell us from where did you see this integral?
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  6. #6
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    Re: Problems trying to solve an integral

    Hi,

    find attached 2 pictures scanned from the book where I saw that integral. As you can see, in the book appears a+rho I put u so, in the book the integral is:

    \displaystyle\int\sqrt[ ]{\frac{(a+rho)^2-x^2}{(a+rho)^2-a^2}}(\frac{a}{a+rho})d(a+rho)

    and I put this one:

    \displaystyle\int\sqrt[ ]{\frac{u^2-x^2}{u^2-a^2}}(\frac{a}{u})du

    but the integral with respect to u should be the same, right?
    Attached Thumbnails Attached Thumbnails Problems trying to solve an integral-integral-1.jpg   Problems trying to solve an integral-integral-2.jpg  
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