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anonymous

  • 5 years ago

how do you integrate (x^9)(cos(x^5))?

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  1. anonymous
    • 5 years ago
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    \[intx ^{9}cosx ^{5}dx. \]

  2. anonymous
    • 5 years ago
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    Possibly integration by parts

  3. anonymous
    • 5 years ago
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    i realize that, but should i don't know what values to assign for u and dv. should i substitute before i even use integration by parts?

  4. anonymous
    • 5 years ago
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    Try x^9 as u

  5. anonymous
    • 5 years ago
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    ((x^10)/10)(-sinx(x^5))

  6. anonymous
    • 5 years ago
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    What is that Nick, like the teacher says, show your work.

  7. anonymous
    • 5 years ago
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    okay Integrate (x^9)we get (x^9+1)/(9+1) intregrate cos(x^5) we get -sin(x^5)

  8. anonymous
    • 5 years ago
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    But there is a rule, the two x quantities are multiplied, so you can't integrate straight out, right?

  9. anonymous
    • 5 years ago
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    yeah, so you integrate by parts.

  10. anonymous
    • 5 years ago
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    yes

  11. anonymous
    • 5 years ago
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    substituting u for x^9 doesn't seem to work chaguanas ...

  12. anonymous
    • 5 years ago
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    Check this I haven't done integration by parts in a while\[x ^{9}\cos x ^{5}-45x ^{5}\sin x ^{5}\]

  13. anonymous
    • 5 years ago
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    +C

  14. watchmath
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=integral+x^9*cos%28x^5%29+dx

  15. watchmath
    • 5 years ago
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    wolframalpha will give you the steps as well.

  16. anonymous
    • 5 years ago
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    hold on are you sure that the integral of cos x^5 is is -sin x^5 - try differentiating -sin x^5.

  17. anonymous
    • 5 years ago
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    Yeah, mine is wrong, see the wolfram watchmath put up. That was the u sub that ecollison was talking about, I missed that.

  18. anonymous
    • 5 years ago
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    watchmath thanks! the link was right

  19. anonymous
    • 5 years ago
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    But please explain the u sub, they didn't account for the x^9 fully.

  20. anonymous
    • 5 years ago
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    u=x^5 not x^9

  21. anonymous
    • 5 years ago
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    lols

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