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anonymous

  • 5 years ago

Solve the integral (sin^2(x)*cos^2(x)) dx / (sin^8(x) - cos^8(x)) .

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  1. anonymous
    • 5 years ago
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    is your dx supposed to be part of the numerator or you typed that wrong?

  2. anonymous
    • 5 years ago
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    it's regular dx, after everything

  3. anonymous
    • 5 years ago
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    so it's supposed to be \[(\sin ^{2}x \cos ^{2}x/\sin ^{8}x-\cos ^{8}x) dx?\]

  4. anonymous
    • 5 years ago
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    6 inches in diameter, 5inches in height, what is the total number of cubic inches. please help what am I doin wrong. 3.14*3squared*5* I don,t know what o do next

  5. anonymous
    • 5 years ago
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    presidential you can't just hijack his thread. start your own please.

  6. anonymous
    • 5 years ago
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    sin^8x-cos^8x is together in the denominator

  7. anonymous
    • 5 years ago
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    This one is kinda complicated so I'd like you to go here http://www.wolframalpha.com/input/?i=integrate+%28sin^2xcos^2x%2F%28sin^8x%E2%88%92cos^8x%29%29dx Hope that helps you out.

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spraguer (Moderator)
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