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Math Help - trig sub

  1. #1
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    trig sub

    \int \frac {48 dx} {x^2 \sqrt{x^2+36}}

    I finally get down to;

    = 48/216 \int \frac {1} {tan^2 t}

    Could you please tell me if that's correct? I doubt it is because there's no indication of a integral of that on my table sheet ...
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  2. #2
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     I = \int \frac{ dx}{ x^2 \sqrt{x^2+1} }<br />
    For this integral, it is good to applt trig. sub. but remind that

     I = \int \frac{ \sec^2(t) dt }{ \tan^2(t) \sqrt{ \sec^2(t) } }

     = \int \frac{ \sec^2(t) dt }{ \tan^2(t)  \sec(t) }

     =  \int \frac{ \sec(t) dt }{ \tan^2(t)  }

     = \int \frac{dt}{ \tan(t) \sin(t) }

     = \int \csc(t) \cot(t)~dt

     = - csc(t) + C

     =  - \frac{1}{ \sin( \tan^{-1}(x))} + C

     = - \frac{1}{ \frac{x}{\sqrt{x^2+1}} } + C

     = -\frac{ \sqrt{x^2+1} }{x} + C

    Of course , it is not the answer to your integral , you need to change a little .

    I would also prefer the substitution  x = \frac{1}{t}

     dx = - \frac{dt}{t^2}

    the integral becomes

     - \int \frac{dt/t^2}{ 1/t^2 (\sqrt{1+t^2}/t)}

     = -\int \frac{tdt}{\sqrt{1+t^2}}

     = - \sqrt{1+t^2} +  C

     = - \frac{ \sqrt{1+x^2}}{x} + C
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  3. #3
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    Thank you very much for your response, but is there another way to solve it without the I = and the non-trig sub?
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  4. #4
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    Quote Originally Posted by Archduke01 View Post
    \int \frac {48 dx} {x^2 \sqrt{x^2+36}}

    I finally get down to;

    = 48/216 \int \frac {1} {tan^2 t}

    Could you please tell me if that's correct? I doubt it is because there's no indication of a integral of that on my table sheet ...
    Did you use x=6tan(t)?

    Draw a right triangle. The leg opposite t is x and the adjacent leg is 6.

    Then the hypotenuse is \sqrt{x^2+36}

    So sec(t)=\frac{\sqrt{x^2+36}}{6}

    \sqrt{x^2+36}=6sec(t)

    x^2=36tan^2(t)

    dx=6sec^2(t)\,dt

    Then \int \frac {48 dx} {x^2 \sqrt{x^2+36}}=48\int\frac{6sec^2(t)\,dt}{(36tan^2  (t))(6sec(t))}

    Simplify and integrate
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