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Jonask

  • 3 years ago

prove intergral

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

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

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

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

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

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

  7. abb0t
    • 3 years ago
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    Did you mutliply by conjugate?

  8. Jonask
    • 3 years ago
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    yes

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

  10. Jonask
    • 3 years ago
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    |dw:1363451580417:dw|

  11. Jonask
    • 3 years ago
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    same thing if you can solve it then you actually proved it

  12. Jonask
    • 3 years ago
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    my problem here is tan pi/2 is undefined

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

  14. Jonask
    • 3 years ago
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    last denominator cos^2x

  15. linshan789
    • 3 years ago
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    |dw:1363445220695:dw|

  16. linshan789
    • 3 years ago
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    Did you get anything like that?

  17. Jonask
    • 3 years ago
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    |dw:1363452681715:dw| i thought thig is +

  18. linshan789
    • 3 years ago
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    Yeah, it should.

  19. linshan789
    • 3 years ago
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    Did you get anything like that?

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

  21. phi
    • 3 years ago
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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

  22. Jonask
    • 3 years ago
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    can you explain how you did it

  23. phi
    • 3 years ago
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    Here are the gory details

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

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

  26. Jonask
    • 3 years ago
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    initially \[\int _0^\pi\]

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

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

  30. Jonask
    • 3 years ago
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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

  31. Jonask
    • 3 years ago
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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

  32. phi
    • 3 years ago
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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

  33. Jonask
    • 3 years ago
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    aha eureka!! thanks guys

  34. phi
    • 3 years ago
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    Here is the problem using the correct limits

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  35. Jonask
    • 3 years ago
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    thanks again for helping sry for the error

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