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Math Help - Trig substitution integration problem

  1. #1
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    Trig substitution integration problem

    1. The problem statement, all variables and given/known data

    integral√(9-x²) / (x²)

    2. Relevant equations

    Just trig substitution

    3. The attempt at a solution

    Ok, for trig sub I did

    u=asinΘ
    x=3sinΘ
    9-x²=9-9sin²=9(1-sin²Θ)
    so putting it into the equation
    x²=9sin²Θ
    dx=3cosΘ
    √9cos²Θ=3cosΘ/x^2
    (3cosΘ/9sin²Θ)*(3cosΘ)=9cos²Θ/9sin²Θ= cot²

    but the integral of cot is ln(sinU) or it can be -ln(cscu)

    I am stuck
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  2. #2
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    Before converting it directly into \cot^{2} \theta, look at it in terms of cos and sin:
    \int \frac{\cos^{2} \theta}{\sin^{2} \theta} d\theta = \int \frac{1-\sin^{2}\theta}{\sin^{2} \theta} d\theta = \int \left( \frac{1}{\sin^{2}\theta }- \frac{\sin^{2} \theta}{\sin^{2} \theta}\right)d\theta

    You can simplify it more and it'll be easier to integrate.
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  3. #3
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    So would it end up

    int(csc²-1)

    cot²=csc²-1

    = int(cot²)

    = ln(sinu)*ln(sinu)

    = ln(x/3) * ln(x/3)

    That still does not look right. What am I missing?
    Last edited by kdizzle711; May 6th 2008 at 02:59 PM.
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  4. #4
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    Recall that \frac{d}{dx} \cot \theta = -\csc^{2} \theta

    So you can see what to do with your integral: \int \left(\csc^{2} \theta - 1\right) d\theta
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  5. #5
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    Is it

    \int \left(\csc^{2} \theta - 1\right) d\theta

    \left(-cot \theta - x\right) d\theta

    -(sqrt(9-x^2)/x) - x
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  6. #6
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    Since you're integrating with respect to \theta, \int 1 d\theta = \theta + C not  x + C.

    So your antiderivative is: -\cot \theta - \theta + C.

    So using your initial sub, you have to find an expression for \theta in terms of x.
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  7. #7
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    So

    x=3sin(theta)
    (x/3)=sin(theta)
    sin^-1(x/3)=theta

    So do I just plug that in to
    -cot(sin^-1(x/3))-(sin^-1(x/3))

    Is that the right answer???
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  8. #8
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    Yep you can simplify it more though, specifically the \cot \left[\sin^{-1} \left(\frac{x}{3}\right) \right] part. Recalling that \cot \theta = \frac{\cos \theta}{\sin \theta}

    \cot \left[\sin^{-1} \left(\frac{x}{3}\right) \right] = \frac{\cos \left[\sin^{-1} \left(\frac{x}{3}\right) \right]}{\sin \left[\sin^{-1} \left(\frac{x}{3}\right) \right]}

    Can you simplify it more? Your solution is correct. Just doesn't look that neat with that \sin^{-1}\left(\frac{x}{3}\right) inside the \cot x.
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