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Jonask

prove intergral

  • one year ago
  • one year ago

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  1. Jonask
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    \[\int\limits _0^1\frac{1}{\alpha-\cos x}dx=\frac{\pi}{\sqrt{\alpha^2-1}}\]

    • one year ago
  2. Jonask
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    i tried \[\tan \frac{x}{2}=u\] i have \[\huge \int\limits \frac{1}{\alpha-\frac{1-t^2}{1+t^2}}\frac{2dt}{1+t^2}=2\huge \int\limits \frac{1}{\alpha-1+(1+\alpha)t^2}dt\]

    • one year ago
  3. Jonask
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    \[\huge \frac{2}{\alpha+1}\int\limits \frac{dt}{(\sqrt{\frac{\alpha-1}{\alpha+1}})^2+t^2}=\frac{2}{\alpha+1}\tan^{-1}(\frac{t}{\sqrt{\frac{\alpha-1}{\alpha+1}}})\]

    • one year ago
  4. Jonask
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    first of all tan pi/2 is undefined,is there anything wrong here?

    • one year ago
  5. Jonask
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    oh something missing the whole thing shud be divided by \[\sqrt{\frac{\alpha-1}{\alpha+1}}\]

    • one year ago
  6. Jonask
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    \[\huge \frac{2}{\sqrt{\alpha^2-1}}\tan^{-1}\frac{\frac{x}{2}}{\sqrt{\frac{\alpha-1}{\alpha+1}}}\]

    • one year ago
  7. abb0t
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    Did you mutliply by conjugate?

    • one year ago
  8. Jonask
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    yes

    • one year ago
  9. linshan789
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    Wait, you're proving. My bad, I thought you were solving using trig sub.

    • one year ago
  10. Jonask
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    |dw:1363451580417:dw|

    • one year ago
  11. Jonask
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    same thing if you can solve it then you actually proved it

    • one year ago
  12. Jonask
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    my problem here is tan pi/2 is undefined

    • one year ago
  13. Jonask
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    METHOD II RATIONALISE AND SEPERATE \[\int\limits \frac{1}{\alpha-\cos x}dx=\int\limits \frac{\alpha+\cos x}{\alpha^2-\cos^2x}dx=\int\limits \frac{\alpha}{\alpha^2-\cos^2x}+\int\limits \frac{\cos x}{\alpha^2-\cos x}\]

    • one year ago
  14. Jonask
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    last denominator cos^2x

    • one year ago
  15. linshan789
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    |dw:1363445220695:dw|

    • one year ago
  16. linshan789
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    Did you get anything like that?

    • one year ago
  17. Jonask
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    |dw:1363452681715:dw| i thought thig is +

    • one year ago
  18. linshan789
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    Yeah, it should.

    • one year ago
  19. linshan789
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    Did you get anything like that?

    • one year ago
  20. Jonask
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    yes thats my first solution there on top instead of z i used t

    • one year ago
  21. phi
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    I got \[ \int\limits _0^1\frac{1}{\alpha-\cos x}dx= \frac{2}{\sqrt{\alpha^2-1}} \tan^{-1}\left(\sqrt{\frac{\alpha+1}{\alpha-1}}\tan\frac{1}{2}\right)\] I don't see the inverse tan giving pi/2 except at alpha->1

    • one year ago
  22. Jonask
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    can you explain how you did it

    • one year ago
  23. phi
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    Here are the gory details

    • one year ago
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  24. Jonask
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    these makes it clear ,thanks but how do we simplify to get the final answer

    • one year ago
  25. Jonask
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    \[\frac{\pi}{\sqrt{\alpha^2-1}}\]

    • one year ago
  26. Jonask
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    initially \[\int _0^\pi\]

    • one year ago
  27. Jonask
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    • one year ago
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  28. Jonask
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    i had typed the wrong boundaries 0 to 1 instead of 0 to pi

    • one year ago
  29. Jonask
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    SO THE PROBLEM IS HERE \[\frac{1}{\sqrt{\alpha^2-1}}\tan^{-1}\left( {\sqrt{\frac{\alpha-1}{\alpha+1}}\tan (x/2)}\right)\]

    • one year ago
  30. Jonask
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    \[\frac{1}{\sqrt{\alpha^2-1}}\tan^{-1}\left( {\sqrt{\frac{\alpha-1}{\alpha+1}}\tan (\pi/2)}\right)\] the part inside is undefined

    • one year ago
  31. Jonask
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    \[\frac{2}{\sqrt{\alpha^2-1}}\tan^{-1}\left( {\sqrt{\frac{\alpha-1}{\alpha+1}}\tan (\pi/2)}\right)\] i mean 2

    • one year ago
  32. phi
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    that makes more sense, because I was going to say the original expression is false. however, for this new limit, you can say tan(x) as x-> pi/2 approaches infinity and the inverse tangent of x--> infinity approaches pi/2

    • one year ago
  33. Jonask
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    aha eureka!! thanks guys

    • one year ago
  34. phi
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    Here is the problem using the correct limits

    • one year ago
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  35. Jonask
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    thanks again for helping sry for the error

    • one year ago
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