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Math Help - Trig Integral Questions

  1. #1
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    Trig Integral Questions

    Can't solve these two. Apparently you're NOT supposed to use a trig substitution for the second one. I don't know how you can for the first since 5 isn't a prime number.

    First one:

    ∫ dx/[x*√(5-x^2)]

    Second one:

    ∫ x^3*√(9-x^2) dx

    Thanks
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  2. #2
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    Can't solve these two. Apparently you're NOT supposed to use a trig substitution for the second one. I don't know how you can for the first since 5 isn't a prime number.

    Second one:

    ∫ x^3*√(9-x^2) dx

    Thanks
    Make the substitution u = \sqrt{9 - x^2} \Rightarrow x^2 = 9 - u^2 and dx = - \frac{\sqrt{9 - x^2}}{x} \, du.
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  3. #3
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    Can't solve these two. Apparently you're NOT supposed to use a trig substitution for the second one. I don't know how you can for the first since 5 isn't a prime number.

    First one:

    ∫ dx/[x*√(5-x^2)]

    [snip]
    Do you have to use a trig substitution?

    Life's much easier when you make the substitution u = \sqrt{5 - x^2} \Rightarrow x^2 = 5 - u^2 and dx = - \frac{\sqrt{5 - x^2}}{x} \, ....
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  4. #4
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    I don't see how the second one works with the substitution.

    I can get to x/√5-x^2 dx but then I'm stuck.

    I pretty much don't understand why x's are still in there if I'm trying to put all u's into the equation.
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  5. #5
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    For ∫ x^3*√(9-x^2) dx

    Mr. Fantastic told you what to substitute. If you take u = \sqrt{9-x^2}.

    du = \frac{-\not{2}x}{\not{2}\sqrt{9-x^2}} dx

    du = -\frac{x}{\sqrt{9-x^2}} dx

    dx = -\frac{\sqrt{9-x^2}}{x} du

    Now replace! you'll see that you can substitute 9-x^2 and x^2 (remember Mr. F said u^2 = 9 - x^2).
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