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anonymous

  • one year ago

please helpppppppp... question in comment

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  1. anonymous
    • one year ago
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    wht do u neep help with

  2. anonymous
    • one year ago
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  3. anonymous
    • one year ago
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    the exact value is

  4. IrishBoy123
    • one year ago
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    re-write the integral using the identity and replacing the cube term

  5. anonymous
    • one year ago
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    unable to do..

  6. IrishBoy123
    • one year ago
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    if A = B - C \[\int B = \int A+C\]

  7. anonymous
    • one year ago
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    the intergal will replace the cube term

  8. anonymous
    • one year ago
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    i need detailed explanation if you dont mind !!!!!!!!

  9. anonymous
    • one year ago
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    im talking bout u need to take 1/2 and replace it with the cube term which is 3

  10. anonymous
    • one year ago
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    you have to integrate first????

  11. anonymous
    • one year ago
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    ik thts wht i was telling u

  12. anonymous
    • one year ago
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    will you help in donig it ?

  13. anonymous
    • one year ago
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    who me

  14. anonymous
    • one year ago
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    @phi

  15. anonymous
    • one year ago
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    @mbma526

  16. anonymous
    • one year ago
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  17. phi
    • one year ago
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    do you know how to integrate cos 3x ?

  18. anonymous
    • one year ago
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    no

  19. phi
    • one year ago
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    do you know how to integrate cos x ?

  20. anonymous
    • one year ago
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    no @phi

  21. phi
    • one year ago
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    do you know the derivative of sin x ?

  22. anonymous
    • one year ago
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    -cosx

  23. phi
    • one year ago
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    just cos x \[ \frac{d}{dx} \sin x = \cos x \] we can "undo the derivative" by integrating \[ \int d \sin x = \int \cos x \ dx \\ \sin x = \int \cos x \ dx \] I'm not sure that is clear, but people memorize this integral \[ \int \cos x \ dx = \sin x + C\]

  24. anonymous
    • one year ago
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    but for \[\cos ^{3} x\] ??

  25. phi
    • one year ago
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    let's first do cos 3x if we let u= 3x and we "work backwards" \[ \frac{d}{dx} \sin u = \cos u \frac{d}{dx} u = \cos 3x \cdot 3\\ \frac{d}{dx} \sin 3x = 3 \cos 3x \]

  26. phi
    • one year ago
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    that is the derivative of sin 3x using the chain rule if we want to integrate cos 3x we need a 3 out front: 3 cos 3x and to compensate , we multiply by 1/3 \[ \frac{1}{3} \int 3 \cos 3x \ dx = \frac{1}{3} \sin 3x \]

  27. phi
    • one year ago
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    yes, there is a power of 3. patience.

  28. phi
    • one year ago
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    hopefully you can integrate cos x and cos 3x (see above) we are given \[ \cos 3x =4 \cos^3 x - 3 \cos x \] can you solve for cos^3 x ?

  29. anonymous
    • one year ago
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    no not been able since 1 hour -___-

  30. phi
    • one year ago
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    if you had 3= 4x + y can you solve for x ?

  31. anonymous
    • one year ago
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    yess x=(3-y)/4

  32. phi
    • one year ago
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    use that same idea to "solve" (which means isolate to one side) cos^3 x start with \[ \cos 3x =4 \cos^3 x - 3 \cos x \]

  33. anonymous
    • one year ago
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    didnt read the question well.. thank you @phi

  34. phi
    • one year ago
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    you should get \[ \cos^3 x = \frac{1}{4}\cos 3x + \frac{3}{4} \cos x \]

  35. anonymous
    • one year ago
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    yesss thank you

  36. phi
    • one year ago
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    now the integral \[ \int \cos^3 x \ dx \] can be written as \[ \frac{1}{4}\int \cos 3x \ dx +\frac{3}{4} \int \cos x \ dx\]

  37. anonymous
    • one year ago
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    yess i can do it further.. thank you

  38. phi
    • one year ago
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    yw

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