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Math Help - integration arcsin rule

  1. #1
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    integration arcsin rule

    \int \frac{7x}{\sqrt{16-x^4}} dx = 7\int \frac{x}{\sqrt{4^2-(x^2)^2}} dx

    The arcsin rule can be used here, right? I'm not sure what to do with the x in the numerator to be able to apply the arcsin rule.
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  2. #2
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    Quote Originally Posted by dataspot View Post
    \int \frac{7x}{\sqrt{16-x^4}} dx = 7\int \frac{x}{\sqrt{4^2-(x^2)^2}} dx

    The arcsin rule can be used here, right? I'm not sure what to do with the x in the numerator to be able to apply the arcsin rule.
    First, you will need to use the substitution of u = x^2 to obtain \frac 72 \int \frac 1{\sqrt{16 - u^2}} \ \mathrm{d}u.

    Then you can apply the arcsin rule which is \int \frac{1}{\sqrt{a^2 - x^2}} \ \mathrm{d}x = \mathrm{arcsin}\left(\frac{x}{a}\right).

    Remember to change the u back to x^2 in the final answer.
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  3. #3
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    Quote Originally Posted by Air View Post
    First, you will need to use the substitution of u = x^2 to obtain \frac 72 \int \frac 1{\sqrt{16 - u^2}} \ \mathrm{d}u.

    Then you can apply the arcsin rule which is \int \frac{1}{\sqrt{a^2 - x^2}} \ \mathrm{d}x = \mathrm{arcsin}\left(\frac{x}{a}\right).

    Remember to change the u back to x^2 in the final answer.
    Got it, thanks!
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