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Math Help - Trigonometric Substitution Problem

  1. #1
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    Trigonometric Substitution Problem

    I'm trying to solve the following integral:
    { stands for integral

    {x^3/(sqrt[x^2 +4]) dx

    This is what I have so far:

    @ stands for pheta

    x=2tan@
    dx=2sec^2@ d@

    x^2+4=4(1+tan^2@)
    Sqrt[x^2+4]=2sec@

    Plugging it back in:

    {8tan^3@2sec^2@/2sec@ d@

    which simplifies to

    8{tan^3@sec@ d@

    I can't seem to solve this. I've tried integration by parts as well as simple sub.

    Where am I going wrong?

    Thank you!!
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  2. #2
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    Quote Originally Posted by Jeffman50 View Post
    I'm trying to solve the following integral:
    { stands for integral

    {x^3/(sqrt[x^2 +4]) dx

    This is what I have so far:

    @ stands for pheta

    x=2tan@
    dx=2sec^2@ d@

    x^2+4=4(1+tan^2@)
    Sqrt[x^2+4]=2sec@

    Plugging it back in:

    {8tan^3@2sec^2@/2sec@ d@

    which simplifies to

    8{tan^3@sec@ d@

    I can't seem to solve this. I've tried integration by parts as well as simple sub.

    Where am I going wrong?

    Thank you!!
    Use a u sub

    \int\frac{x^3}{\sqrt{x^2+4}}

    Let u=x^2+4 \iff x^2=u-4 \implies du=2xdx

    subbing into the above integral gives

    \frac{1}{2}\int \frac{u-4}{\sqrt{u}}du=\frac{1}{2}\int (u^{\frac{1}{2}}-4u^{-\frac{1}{2}})du
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  3. #3
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    Interesting.. Is there a way to do it using trigonometric substitution? Its under that heading in my textbook
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  4. #4
    Member Black's Avatar
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    Alternatively,

    8\int \text{tan}^3\theta \, \text{sec}\theta \, d\theta=8\int \text{tan}^2\theta \, \text{sec}\theta\text{tan}\theta \, d\theta = 8\int \left(\text{sec}^2\theta-1\right)\text{sec}\theta\text{tan}\theta \, d\theta.

    Let u=\text{sec}\theta .
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  5. #5
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    Quote Originally Posted by Jeffman50 View Post
    I'm trying to solve the following integral:
    { stands for integral

    {x^3/(sqrt[x^2 +4]) dx

    This is what I have so far:

    @ stands for pheta

    x=2tan@
    dx=2sec^2@ d@

    x^2+4=4(1+tan^2@)
    Sqrt[x^2+4]=2sec@

    Plugging it back in:

    {8tan^3@2sec^2@/2sec@ d@

    which simplifies to

    8{tan^3@sec@ d@

    I can't seem to solve this. I've tried integration by parts as well as simple sub.

    Where am I going wrong?

    Thank you!!
    You're doing great.

    \frac{d}{d\theta}(\sec\theta)=\sec\theta\tan\theta

    \tan^2\theta=\sec^2\theta-1

    Rewrite 8\tan^3\theta\sec\theta\,d\theta as

    8\tan^2\theta\sec\theta\tan\theta\,d\theta

    Good luck
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  6. #6
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    Quote Originally Posted by TheEmptySet View Post
    Use a u sub

    \int\frac{x^3}{\sqrt{x^2+4}}

    Let u=x^2+4 \iff x^2=u-4 \implies du=2xdx

    subbing into the above integral gives

    \frac{1}{2}\int \frac{u-4}{\sqrt{u}}du=\frac{1}{2}\int (u^{\frac{1}{2}}-4u^{-\frac{1}{2}})du
    Your numerator is x^3, not x^2.


    Trigonometric substitution is possible, but difficult.

    It's easier to use Hyperbolic Substitution in this case.

    \int{\frac{x^3}{\sqrt{x^2 + 4}}\,dx}.

    Make the substitution x = 2\sinh{t} so that dx = 2\cosh{t}\,dt.

    Note that t = \sinh^{-1}{\frac{x}{2}}.


    So the integral becomes

    \int{\frac{(2\sinh{t})^3}{\sqrt{(2\sinh{t})^2 + 4}}\,2\cosh{t}\,dt}

     = \int{\frac{16\sinh^3{t}\cosh{t}}{\sqrt{4\sinh^2{t} + 4}}\,dt}

     = \int{\frac{16\sinh^3{t}\cosh{t}}{\sqrt{4(\sinh^2{t  } + 1)}}\,dt}

     = \int{\frac{16\sinh^3{t}\cosh{t}}{2\sqrt{\cosh^2{t}  }}\,dt}

     = \int{\frac{8\sinh^3{t}\cosh{t}}{\cosh{t}}\,dt}

     = 8\int{\sinh^3{t}\,dt}

     = 8\int{\sinh{t}\sinh^2{t}\,dt}

     = 8\int{\sinh{t}(\cosh^2{t} - 1)\,dt}.

    Now let u = \cosh{t} so that du = \sinh{t}\,dt.

    The integral becomes

    8\int{u^2 - 1\,du}

     = 8\left[\frac{1}{3}u^3 - u\right] + C

     = 8\left[\frac{1}{3}\cosh^3{t} - \cosh{t}\right] + C

     = 8\left[\frac{1}{3}\sqrt{(1 + \sinh^2{t})^3} - \sqrt{1 + \sinh^2{t}}\right] + C

     = 8\left\{\frac{1}{3}\left[\sqrt{1 + \left(\frac{x}{2}\right)^2}\right]^3 - \sqrt{1 + \left(\frac{x}{2}\right)^2}\right\} + C

     = 8\left\{\frac{1}{3}\left[\frac{\sqrt{4 + x^2}}{2}\right]^3 - \frac{\sqrt{4 + x^2}}{2}\right\} + C

     = 8\left[\frac{\sqrt{(4 + x^2)^3}}{24} - \frac{\sqrt{4 + x^2}}{2}\right] + C

     = \frac{\sqrt{(4 + x^2)^3}}{3} - 4\sqrt{4 + x^2} + C
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