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anonymous

  • 5 years ago

what is integral of (cos x) / x ?

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  1. anonymous
    • 5 years ago
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    What integration techniques do you have right now?

  2. anonymous
    • 5 years ago
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    techniques? like?

  3. anonymous
    • 5 years ago
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    u-substitution parts trig substitution etc.

  4. anonymous
    • 5 years ago
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    i just wanna find out derivative of cosx/x between 3 pi/2 and pi/2

  5. anonymous
    • 5 years ago
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    and then the integral of cos x/x

  6. anonymous
    • 5 years ago
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    The derivative would be the slope of the tangent line, the integral is the area of the region. Two different (yet related) things.

  7. anonymous
    • 5 years ago
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    i have tried finding out integral of cos x/x but it was an infinite loop!!! need an answer for this plz

  8. anonymous
    • 5 years ago
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    i have tried finding out integral of cos x/x but it was an infinite loop!!! need an answer for this plz

  9. anonymous
    • 5 years ago
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    ntegration by parts is a heuristic rather than a purely mechanical process for solving integrals; given a single function to integrate, the typical strategy is to carefully separate it into a product of two functions ƒ(x)g(x) such that the integral produced by the integration by parts formula is easier to evaluate than the original one.

  10. anonymous
    • 5 years ago
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    So it sounds like you tried parts. Which is what I would do. The integral ends up being: \[[\ln(x)\cos(x)+\sin(x)/x]/2 + c\]

  11. anonymous
    • 5 years ago
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    uh huh.

  12. anonymous
    • 5 years ago
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    how did u get that answer? can u plz explain the steps..

  13. anonymous
    • 5 years ago
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    I set u = cos(x) and dv = dx/x and solved for du and v. Then I put in uv - ∫v du and had to use parts again using u = sin(x) and dv = ln(x)dx and solved for du and v again. This time I ended up with ln(x)cos(x)+sin(x)/x - ∫cos(x)/x dx = ∫cos(x)/x dx (the original problem) then I added ∫cos(x)/x dx to both sides and divide by two.

  14. anonymous
    • 5 years ago
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    hmm..so derv of cos x =-sinx and intg(dv) =1/x dx right?

  15. anonymous
    • 5 years ago
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    du = -sin(x)dx and v = ln(x) for the first parts.

  16. anonymous
    • 5 years ago
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    the formula is , u int (v)- int{d/dx(u) . int(v)}dx right?

  17. anonymous
    • 5 years ago
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    ∫u dv = u*v - ∫v du is what integration by parts states.

  18. anonymous
    • 5 years ago
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    intr(u) dv or integral (uv)?

  19. anonymous
    • 5 years ago
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    dont we ve any formula as "something / somethig" in intergral as we ve for derivatives?

  20. anonymous
    • 5 years ago
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    This integral cannot be represented in terms of transcendental and algebraic functions. You can only represent it with series.

  21. anonymous
    • 5 years ago
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    if we have lower limit as pi/2 and upper one as 3 pi/2 then?

  22. anonymous
    • 5 years ago
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    as the sin terms become all zeroes with pi/2 's

  23. anonymous
    • 5 years ago
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    It can't be done with any method other than series, even if you constrained the limits.

  24. anonymous
    • 5 years ago
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    im sry the cos terms

  25. anonymous
    • 5 years ago
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    ohh..isnt there any other way out?

  26. anonymous
    • 5 years ago
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    Not from what I researched online. You may be able to use power series. However, u-substitution, backwards u-substitution, integration by parts, partial fractions, etc. will not solve this integral. If I could remember power series from three years ago I could help, but I don't remember :(

  27. anonymous
    • 5 years ago
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    ohh :(...anyways thanks

  28. anonymous
    • 5 years ago
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    gotta findout frm somewhere else then

  29. anonymous
    • 5 years ago
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    thanks Tbates and jshowa

  30. anonymous
    • 5 years ago
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    hey can u just temme how do i join the group chat here??

  31. anonymous
    • 5 years ago
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    i have got lot many questions to ask....!!!

  32. anonymous
    • 5 years ago
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    anybody there?????

  33. anonymous
    • 5 years ago
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    I'm not sure what you do, I tried typing something and it didn't work for me. You can just ask another question.

  34. anonymous
    • 5 years ago
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    hmm...

  35. anonymous
    • 5 years ago
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    so in such a case we get loops isnt it?

  36. anonymous
    • 5 years ago
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    even if got the upper n lower limits for that

  37. anonymous
    • 5 years ago
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    can anyone answer this?

  38. anonymous
    • 5 years ago
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    I'm not sure, however, I intend on answering this question once I refresh myself with series. So, check back in a day or so. The limits really don't matter because you still have to evaluate the integral before you use them. And with integration by parts, you will continually get integrals that must be done with parts and it will never end. You must either construct a power series or Taylor series of cosx/x and estimate the integral.

  39. anonymous
    • 5 years ago
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    yeah..that actually does go on n on..

  40. anonymous
    • 5 years ago
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    n theres one more thing

  41. anonymous
    • 5 years ago
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    n theres one more thing

  42. anonymous
    • 5 years ago
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    tangent is a plane and normal is a line on that plane..is this staement correct?

  43. anonymous
    • 5 years ago
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    anyone there?

  44. anonymous
    • 5 years ago
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    integration by parts would do it

  45. anonymous
    • 5 years ago
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    make 1/x your dv and and your u cos(x)

  46. anonymous
    • 5 years ago
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    hmm...and how wud the final answer look like?

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