Quantcast

A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Jonask

  • 2 years ago

prove intergral

  • This Question is Closed
  1. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  2. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\huge \frac{2}{\alpha+1}\int\limits \frac{dt}{(\sqrt{\frac{\alpha-1}{\alpha+1}})^2+t^2}=\frac{2}{\alpha+1}\tan^{-1}(\frac{t}{\sqrt{\frac{\alpha-1}{\alpha+1}}})\]

  4. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    first of all tan pi/2 is undefined,is there anything wrong here?

  5. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh something missing the whole thing shud be divided by \[\sqrt{\frac{\alpha-1}{\alpha+1}}\]

  6. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\huge \frac{2}{\sqrt{\alpha^2-1}}\tan^{-1}\frac{\frac{x}{2}}{\sqrt{\frac{\alpha-1}{\alpha+1}}}\]

  7. abb0t
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Did you mutliply by conjugate?

  8. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes

  9. linshan789
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Wait, you're proving. My bad, I thought you were solving using trig sub.

  10. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1363451580417:dw|

  11. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    same thing if you can solve it then you actually proved it

  12. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    my problem here is tan pi/2 is undefined

  13. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    last denominator cos^2x

  15. linshan789
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1363445220695:dw|

  16. linshan789
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Did you get anything like that?

  17. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1363452681715:dw| i thought thig is +

  18. linshan789
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yeah, it should.

  19. linshan789
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Did you get anything like that?

  20. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes thats my first solution there on top instead of z i used t

  21. phi
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can you explain how you did it

  23. phi
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Here are the gory details

    1 Attachment
  24. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    these makes it clear ,thanks but how do we simplify to get the final answer

  25. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\frac{\pi}{\sqrt{\alpha^2-1}}\]

  26. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    initially \[\int _0^\pi\]

  27. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1 Attachment
  28. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i had typed the wrong boundaries 0 to 1 instead of 0 to pi

  29. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\frac{1}{\sqrt{\alpha^2-1}}\tan^{-1}\left( {\sqrt{\frac{\alpha-1}{\alpha+1}}\tan (\pi/2)}\right)\] the part inside is undefined

  31. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\frac{2}{\sqrt{\alpha^2-1}}\tan^{-1}\left( {\sqrt{\frac{\alpha-1}{\alpha+1}}\tan (\pi/2)}\right)\] i mean 2

  32. phi
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    aha eureka!! thanks guys

  34. phi
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Here is the problem using the correct limits

    1 Attachment
  35. Jonask
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks again for helping sry for the error

  36. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.