Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

precal

Integration problem

  • one year ago
  • one year ago

  • This Question is Closed
  1. precal
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1349015981615:dw|

    • one year ago
  2. bahrom7893
    Best Response
    You've already chosen the best response.
    Medals 0

    This looks like one of the trig substitutions.

    • one year ago
  3. bahrom7893
    Best Response
    You've already chosen the best response.
    Medals 0

    or maybe integration by parts would work.

    • one year ago
  4. precal
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1349016003653:dw|

    • one year ago
  5. precal
    Best Response
    You've already chosen the best response.
    Medals 0

    so I will rewrite it as|dw:1349016056313:dw|

    • one year ago
  6. precal
    Best Response
    You've already chosen the best response.
    Medals 0

    Am I doing it correct so far?

    • one year ago
  7. bahrom7893
    Best Response
    You've already chosen the best response.
    Medals 0

    I think it's best if we leave it inside. Something tells me that will have to stay as u^(3/2)

    • one year ago
  8. precal
    Best Response
    You've already chosen the best response.
    Medals 0

    then I don't see with trig sub I am suppose to use

    • one year ago
  9. bahrom7893
    Best Response
    You've already chosen the best response.
    Medals 0

    oh nvm

    • one year ago
  10. justine027
    Best Response
    You've already chosen the best response.
    Medals 0

    Nope . Use Inverse sec :) or Sine? I forgot what to use but simple as that :D

    • one year ago
  11. bahrom7893
    Best Response
    You've already chosen the best response.
    Medals 0

    It is trig, u-sub won't work. Yeah keep it out, it's inverse sin of 6x i believe

    • one year ago
  12. precal
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1349016200316:dw|

    • one year ago
  13. justine027
    Best Response
    You've already chosen the best response.
    Medals 0

    Its nerp it is sin inverse x/6 +C :)

    • one year ago
  14. precal
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1349016274224:dw| is what my solution is suppose to be. Of course I am more interested in how to get to the solution

    • one year ago
  15. Callisto
    Best Response
    You've already chosen the best response.
    Medals 5

    I think we should use sec instead of sin?! Let x = 6sec (theta)

    • one year ago
  16. bahrom7893
    Best Response
    You've already chosen the best response.
    Medals 0

    36Sec^2(theta) - 36 = 36 tan^2(theta)

    • one year ago
  17. bahrom7893
    Best Response
    You've already chosen the best response.
    Medals 0

    so the bottom is: sqrt (36 tan^2(theta)) = 6tan^2(theta), yea callisto's right

    • one year ago
  18. bahrom7893
    Best Response
    You've already chosen the best response.
    Medals 0

    and top is: dx = 6Sec(theta)Tan(theta)dtheta

    • one year ago
  19. precal
    Best Response
    You've already chosen the best response.
    Medals 0

    so is my given solution incorrect?

    • one year ago
  20. bahrom7893
    Best Response
    You've already chosen the best response.
    Medals 0

    yea precal

    • one year ago
  21. precal
    Best Response
    You've already chosen the best response.
    Medals 0

    I have some integrals at the backof my book "Integral involving x^2-a^2; a>0"

    • one year ago
  22. siddhantsharan
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\int\limits \frac{ 6secutanudu }{ (36\sec^2u - 36)^{3/2} } = \int\limits \frac{ 6secu tanu du }{ 6^3 \tan^3u }\] \[1/36 \int\limits secdu/\tan^2u \] \[1/36 \int\limits \frac{ cosudu }{ \sin^2u? }\]

    • one year ago
  23. Callisto
    Best Response
    You've already chosen the best response.
    Medals 5

    cos u/ sin^2 u = cscu cotu \[\int \csc u \cot udu = -cscu +C\] Time to draw a triangle...

    • one year ago
  24. precal
    Best Response
    You've already chosen the best response.
    Medals 0

    time to draw a triangle? how is that going to help....I wonder if we are taking the correct approach. I was given problems to study for an upcoming exam. Are we on the correct path?

    • one year ago
  25. precal
    Best Response
    You've already chosen the best response.
    Medals 0

    I don't see how this is going to get me the solution I was given at all????

    • one year ago
  26. siddhantsharan
    Best Response
    You've already chosen the best response.
    Medals 1

    Take sinu as t. => -dt = cosudu \[1/36 \int\limits \frac{ -dt }{ t^2 } = \frac{ 1 }{ 36t } + C = cosecu/36 + C = \frac{ \sqrt6 }{ 36\sqrt{x^2 - 6} } \]

    • one year ago
  27. precal
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1349017386008:dw|

    • one year ago
  28. alanli123
    Best Response
    You've already chosen the best response.
    Medals 0

    LOL ITS FUNNY CUZ UR NAME IS PRECAL AND UR DOING CAL LOL

    • one year ago
  29. Callisto
    Best Response
    You've already chosen the best response.
    Medals 5

    |dw:1349017310113:dw| since we let x = 6 secu secu = x/6 cosu = 6/x You can find the opposite side, hence sin u, and cscu in terms of x

    • one year ago
  30. precal
    Best Response
    You've already chosen the best response.
    Medals 0

    gotta study all types of math. I am just more comfortable at the precal and below level

    • one year ago
  31. siddhantsharan
    Best Response
    You've already chosen the best response.
    Medals 1

    Yeaah my bad. Thats \[\sqrt x\] NOT \[\sqrt 6\]

    • one year ago
  32. precal
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1349017504869:dw|

    • one year ago
  33. Callisto
    Best Response
    You've already chosen the best response.
    Medals 5

    Yes. And what is sin u and cscu?

    • one year ago
  34. precal
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1349017612360:dw|

    • one year ago
  35. Callisto
    Best Response
    You've already chosen the best response.
    Medals 5

    And cscu?

    • one year ago
  36. precal
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1349017652671:dw|

    • one year ago
  37. Callisto
    Best Response
    You've already chosen the best response.
    Medals 5

    And remember we got \[-\frac{1}{36}cscu+C\]For the integral?

    • one year ago
  38. precal
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1349017768354:dw|

    • one year ago
  39. siddhantsharan
    Best Response
    You've already chosen the best response.
    Medals 1

    ALTERNATE: \[x^2 - 36 = t^2 =>tdt=xdx\] \[I = \int\limits \frac{ tdt }{ t^3\sqrt{t^2 + 36}} = \int\limits \frac{ dt }{t^3 \sqrt{ 1 + \frac{ 36 }{ t^2 }} }\]

    • one year ago
  40. precal
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks Callisto, you were correct (back to the basics)

    • one year ago
  41. Callisto
    Best Response
    You've already chosen the best response.
    Medals 5

    You're welcome :)

    • one year ago
  42. precal
    Best Response
    You've already chosen the best response.
    Medals 0

    that is in reference to drawing the triangle. Yes, I got my solution but now I have to go back and study the process

    • one year ago
  43. siddhantsharan
    Best Response
    You've already chosen the best response.
    Medals 1

    Take: \[ 1 + \frac{ 36 }{ t^2 } = u = > du = -36dt/t^3\]

    • one year ago
  44. siddhantsharan
    Best Response
    You've already chosen the best response.
    Medals 1

    You would get the same answer again after the sub. :)

    • one year ago
  45. precal
    Best Response
    You've already chosen the best response.
    Medals 0

    @siddhantsharan I am always interested in the alternate approach, especially if it gets me to the answer quicker

    • one year ago
  46. precal
    Best Response
    You've already chosen the best response.
    Medals 0

    thanks everyone, gotta take a break and look at this later........

    • one year ago
  47. Callisto
    Best Response
    You've already chosen the best response.
    Medals 5

    Perhaps it's easier for you to read. (PS: I'm not giving direct answer!!) First Method: By Trigo Sub.\[\int \frac{dx}{(x^2 -36)^\frac{3}{2}}\]Let x=6sec u; dx = 6secu tanu du. Then, the integral becomes\[=\int \frac{6secu \ tanu\ du}{((6secu)^2 -36)^\frac{3}{2}}\]\[=\int \frac{6secu \ tanu\ du}{[36(sec^2u -1)]^\frac{3}{2}}\]\[=\int \frac{6secu \ tanu\ du}{[36(tan^2u)]^\frac{3}{2}}\]\[=\frac{1}{36}\int \frac{secu \ tanu\ du}{tan^3u}\]\[=\frac{1}{36}\int cscutanudu\]\[=-\frac{1}{36}\csc u+C\] Since x=6secu , secu = x/6 , cosu = 6/x By Pyth. Thm., opposite side is \(\sqrt{x^2-6^2}=\sqrt{x^2-36}\) So, cscu = \(\frac{x}{\sqrt{x^2-36}}\). Substituting this into the last step, you can get your answer.

    • one year ago
  48. Callisto
    Best Response
    You've already chosen the best response.
    Medals 5

    Second method: by substitution, suggested by @siddhantsharan \[\int \frac{dx}{(x^2 -36)^\frac{3}{2}}\]Let x^2-36 = t^2 ; 2x dx = 2tdt => x dx = t dt => \(dx = \frac{t dt}{ x} = \frac{t dt}{ \sqrt{t^2+36}}\) PS: we already know x \(\ge\) 6, so it should be +ve sqrt The integral becomes \[\int \frac{tdt}{\sqrt{(t^2 +36)}(t^3)}\]\[=\int \frac{\frac{tdt}{t}}{\sqrt{\frac{(t^2 +36)}{t^2}}(t^3)}\] \[=\int \frac{dt}{\sqrt{1+\frac{36}{t^2}}(t^3)}\]Let u = 1+ (36/t^3) ; du = \(\frac{-72}{t^3}\)dt The integral becomes \[=-\frac{1}{72}\int \frac{du}{\sqrt{u}}\]\[=-\frac{1}{72}(2\sqrt{u}) +C\]\[=-\frac{1}{36}(\sqrt{u}) +C\] u= 1+ (36/t^2) And t^2 = x^2-36, So,\( u = 1 +\frac{36}{x^2-36} = \frac{x^2}{x^2-36}\) Sub. \( u = \frac{x^2}{x^2-36}\) into the last step, and simplify it, you should be able to get the answer. It's so amazing :D

    • one year ago
  49. siddhantsharan
    Best Response
    You've already chosen the best response.
    Medals 1

    @Callisto Clarified succinctly. Really well done :D

    • one year ago
  50. precal
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks you are so awesome :) Both of you, wish I could give more than 1 medal

    • one year ago
    • Attachments:

See more questions >>>

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.