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anonymous

  • 5 years ago

Trignometric substitutio of the indefinite integral x^3*sqrt(16+x^2)dx

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  1. Owlfred
    • 5 years ago
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    Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

  2. myininaya
    • 5 years ago
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    just a sec k?

  3. anonymous
    • 5 years ago
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    k

  4. myininaya
    • 5 years ago
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    so far this is what i have

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  5. anonymous
    • 5 years ago
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    kk looks good so far

  6. myininaya
    • 5 years ago
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    http://answers.yahoo.com/question/index?qid=20080508091701AAA0Mq0 im fixing to have dinner this link should help you with the rest im sorry

  7. anonymous
    • 5 years ago
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    the answer should be [(16-x^2)^5/2]/5-[16(16+x^2)^3/2]/3+C

  8. myininaya
    • 5 years ago
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    you should also put your answer back in terms of x i might be back on later if you haven't figured it out by then k?

  9. anonymous
    • 5 years ago
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    k thanks for your help.

  10. myininaya
    • 5 years ago
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    im sorry hai peace

  11. anonymous
    • 5 years ago
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    can you please help me?

  12. anonymous
    • 5 years ago
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    amistre its really tough

  13. amistre64
    • 5 years ago
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    x^3*sqrt(16+x^2) dx; yeah, it can be..

  14. amistre64
    • 5 years ago
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    4tan(t) = x^2

  15. amistre64
    • 5 years ago
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    ack!!.. 4tan(t) = x

  16. amistre64
    • 5 years ago
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    64 tan^3(t) * 4 sec(t) dt ; just gotta determine the dt now

  17. anonymous
    • 5 years ago
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    its 4sec^2 theta dtheta

  18. amistre64
    • 5 years ago
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    4sec^2(t) dt = dx right?

  19. anonymous
    • 5 years ago
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    yea

  20. amistre64
    • 5 years ago
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    64 tan^3(t) * 16sec^3(t) dt then right?

  21. anonymous
    • 5 years ago
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    yea

  22. anonymous
    • 5 years ago
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    i put the answer above for reference, this problem is driving me crzy

  23. amistre64
    • 5 years ago
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    1024 (tan(t)sec(t))^3

  24. anonymous
    • 5 years ago
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    can you do it on your own and see if you get that?

  25. amistre64
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=int%2864+tan^3%28t%29+*+16sec^3%28t%29%29+dt

  26. amistre64
    • 5 years ago
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    I aint got the time tonight, libraries closing soon but wolfram gives steps

  27. amistre64
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=int%28x^3+sqrt%2816%2Bx^2%29%29dx or this for the original

  28. anonymous
    • 5 years ago
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    You've got it. You've worked your way to a function that can be integrated.

  29. anonymous
    • 5 years ago
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    Integral (sec theta tan theta)^3=sec^3 theta. Now you have to convert back. Good job Amistre. He just had to run and didn't have time to interpret it.

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