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Math Help - Rationalizing Substitutions

  1. #1
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    Post Rationalizing Substitutions

    [IMG]file:///C:/Users/Karina/AppData/Local/Temp/msohtmlclip1/01/clip_image002.gif[/IMG]Rationalizing Substitutions-math.png


    This is the equation I am having trouble with. I think that 2sin(u) is what I need to use for x. Once I substitute everything in and begin to do all the steps, I get stuck. I'll show what I have so far. If I done something wrong please feel free to correct me. I also know what my answer should be, but I don't want just the answer I need help understanding how to get the answer.


    Rationalizing Substitutions-math-2.png

    [IMG]file:///C:/Users/Karina/AppData/Local/Temp/msohtmlclip1/01/clip_image006.gif[/IMG]
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  2. #2
    Senior Member apcalculus's Avatar
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    Is WolframAlpha acceptable? If it is, be sure to click on 'Show Steps'.


    http://www.wolframalpha.com/input/?i=integrate+(sqrt(4-x^2)/x)

    Good luck!
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  3. #3
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    You got to

    2\int{\frac{\cos^2{u}}{\sin{u}}\,du} alright.

     = 2\int{\frac{1 - \sin^2{u}}{\sin{u}}\,du}

     = 2\int{\frac{1}{\sin{u}} - \frac{\sin^2{u}}{\sin{u}}\,du}

     = 2\int{\csc{u} - \sin{u}\,du}

     = 2\left[-\ln{(\cot{u} + \csc{u})} + \cos{u} + C\right].


    Now try to convert it back to a function of x.
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  4. #4
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    Can you use the half angle formula for cos^2x? Or would that make the problem harder to evaluate? cos^2x = (1+cos^2x)/2
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  5. #5
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    First of all

    \cos^2{x} does NOT equal \frac{1 + \cos^2{x}}{2}, it's actually \frac{1 + \cos{2x}}{2}.

    Second, no you can't. Since you're trying to simplify the integrand, you would want the numerator to be something with the sine function, so that you can cancel.
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