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  1. #1
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    integral

    what is the integral of 1/\sqrt (36-16x^2)

    use substitution
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  2. #2
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    Quote Originally Posted by kasiaogor View Post
    what is the integral of 1/\sqrt (36-16x^2)

    use substitution
    \displaystyle \frac{1}{\sqrt{36-16x^2}}=\frac{1}{4}\frac{1}{\sqrt{\left( \frac{6}{4}\right)^2-x^2}}= \frac{1}{4}\frac{1}{\sqrt{\left( \frac{3}{2}\right)^2-x^2}}

    So the question you need to ask yourself is what trig sub works for things of the form

    \displaystyle \frac{1}{\sqrt{a^2-x^2}}
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