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Math Help - Trigonometric Substitution Integral

  1. #1
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    Trigonometric Substitution Integral

    I'm having trouble on one specific step of this integral:

    Original Equation:  \int \frac{-5x}{(x^2+5)^\frac{3}{2}} dx

    where  x = \sqrt{5} tan\Theta
    dx = \sqrt{5} sec^2\Theta d\Theta

    = \int \frac{-5x}{((\sqrt{5} tan\Theta)^2+5)^\frac{3}{2}} \sqrt{5} sec^2\Theta d\Theta


    Now I don't understand how to get to the next step of:

     -\sqrt{5} \int \frac{tan \Theta}{sec \Theta} d\Theta

    because I get here and I know it's not right:

     -5 \int \frac{\sqrt{5} tan\Theta}{ \sqrt{125}\sqrt{sec^6\Theta}} \sqrt{5}sec^2\Theta d\Theta

    Thanks for the help in advance.
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  2. #2
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    Quote Originally Posted by cynlix View Post
    I'm having trouble on one specific step of this integral:

    Original Equation:  \int \frac{-5x}{(x^2+5)^\frac{3}{2}} dx

    where  x = \sqrt{5} tan\Theta
    dx = \sqrt{5} sec^2\Theta d\Theta

    = \int \frac{-5x}{((\sqrt{5} tan\Theta)^2+5)^\frac{3}{2}} \sqrt{5} sec^2\Theta d\Theta


    Now I don't understand how to get to the next step of:

     -\sqrt{5} \int \frac{tan \Theta}{sec \Theta} d\Theta

    because I get here and I know it's not right:

     -5 \int \frac{\sqrt{5} tan\Theta}{ \sqrt{125}\sqrt{sec^6\Theta}} \sqrt{5}sec^2\Theta d\Theta

    Thanks for the help in advance.
    You can just use a u substitution.

    Let u = x^2 + 5 so that \frac{du}{dx} = 2x.

    The integral becomes:

    \int{\frac{-5x}{(x^2 + 5)^{\frac{3}{2}}}\,dx} = -\frac{5}{2}\int{\frac{2x}{(x^2 + 5)^{\frac{3}{2}}}\,dx}

     = -\frac{5}{2}\int{\frac{1}{u^{\frac{3}{2}}}\,\frac{d  u}{dx}\,dx}

     = -\frac{5}{2}\int{u^{-\frac{3}{2}}\,du}.


    You should be able to go from here...
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  3. #3
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    thank you for that insight, however I need to use the trigonometric substitution for this homework problem.
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  4. #4
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    Quote Originally Posted by cynlix View Post
    I'm having trouble on one specific step of this integral:

    Original Equation:  \int \frac{-5x}{(x^2+5)^\frac{3}{2}} dx

    where  x = \sqrt{5} tan\Theta
    dx = \sqrt{5} sec^2\Theta d\Theta

    = \int \frac{-5x}{((\sqrt{5} tan\Theta)^2+5)^\frac{3}{2}} \sqrt{5} sec^2\Theta d\Theta


    Now I don't understand how to get to the next step of:

     -\sqrt{5} \int \frac{tan \Theta}{sec \Theta} d\Theta

    because I get here and I know it's not right:

     -5 \int \frac{\sqrt{5} tan\Theta}{ \sqrt{125}\sqrt{sec^6\Theta}} \sqrt{5}sec^2\Theta d\Theta

    Thanks for the help in advance.
    You also need to replace the x in your numerator with \sqrt{5}\tan{\theta}.
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