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anonymous

  • 5 years ago

proove by integraion that (1/2)arctan(2x/1-2X) and arctan(1/x) are both integrals of the same function

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  1. anonymous
    • 5 years ago
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    think u mean by differentiaion

  2. anonymous
    • 5 years ago
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    ok help me out please

  3. anonymous
    • 5 years ago
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    no it cant be by differentiation as we r tryin to roll back the two integrals but ignore the c in F(x) + \[c \int\limits_{?}^{?}\]

  4. anonymous
    • 5 years ago
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    F(x) + c

  5. anonymous
    • 5 years ago
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    so the first is \[\frac{1}{2}\tan^{-1} (\frac{2x}{1-2x})\]

  6. anonymous
    • 5 years ago
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    dont think its possbile they must only differ by a constant

  7. amistre64
    • 5 years ago
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    [S] [tan^-1 (2x/1-2X)]/2 = [S] tan^-1(1/x) right?

  8. anonymous
    • 5 years ago
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    prob try some sum of angle formulas, possibly, but I dnt think it will make a difference

  9. anonymous
    • 5 years ago
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    thanx guys.....yes they have to differ by a constant.....its time consuming whew......how can i master induction of inequalities?

  10. anonymous
    • 5 years ago
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    huh, maybe diff of angles formula

  11. amistre64
    • 5 years ago
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    tan^-1(1/x) = u -1/x^2 --------- dx = du perhaps 1 + 1/x^2

  12. amistre64
    • 5 years ago
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    tan(u) = 1/x is what i was thinking lol; lost track mid type

  13. anonymous
    • 5 years ago
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    the difference is an angle of \[\pi/2\]

  14. anonymous
    • 5 years ago
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    Are you sure you've stated the problem correctly? If you meant \[1/2\arctan(2x/1-x^2)\] ie x^2 rather than 2x on the bottom line, then the two expressions are identical, because tan2x = 2tan(x/2)/(1+tan^2(x/2)

  15. anonymous
    • 5 years ago
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    eash man my bad ey guess you are a geenie

  16. anonymous
    • 5 years ago
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    what, \[\frac{1}{2}\tan^{-1} (\frac{2x}{1-x^2}) \neq \tan^{-1} (\frac{1}{x})\]

  17. anonymous
    • 5 years ago
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    You are quite right. Again! I'm not there yet, but using brute force with Mathematica might give us some ideas. I checked 2x on the bottom line as well as x^2: Integrate[ArcTan[1/x], x] == x*ArcCot[x] + Log[1 + x^2]/2 Integrate 1/2tan^-1(2x/(1-x^2) == (x*ArcTan[(2*x)/(1 - x^2)] - Log[1 + x^2])/2 Integrate 1/2tan^-1(2x/(1-2x) == (ArcTan[1 - 4*x]/4 + x*ArcTan[(2*x)/(1 - 2*x)] - Log[1 - 4*x + 8*x^2]/8)/2

  18. anonymous
    • 5 years ago
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    ^ I wouldnt worry about thinking about it, it was a bad question to begin with, I am very sure there was a problem with it

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