A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

whats the integral of sqrt 1+ x^2

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

    You have to make a substitution. Set x=sinh(theta) and take it from there.

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

    \[\int\limits_{}{}\sqrt{1+x^2}dx=\int\limits_{}{}\sqrt{1=\sinh^2 \theta}.\cosh \theta d \theta=\int\limits_{}{}\cosh^2 \theta d \theta\]

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

    1+sinh^2 I mean

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

    \[\cosh^2 \theta = \frac{\cosh 2 \theta +1 }{2}\]where I've used\[\cosh^2 \theta + \sinh^2 \theta = \cosh2 \theta\]and \[\cosh^2 \theta - \sinh^2 \theta =1\]

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

    should we use tangent bc there is a plus sign

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

    Then your integral is,\[\frac{1}{2}\int\limits_{}{}(\cosh2 \theta +1) d \theta=\frac{1}{2}[\frac{1}{2}\sinh 2 \theta + \theta]+c\]

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

    You can use tangent if you want. That's another way of doing it. I just find hyperbolic trigs easier.

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

    Is that your preferred method?

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

    yah

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

    Okay...lol...then set x=tan(theta) and go...

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

    Your integral would be, \[\int\limits_{}{}\sqrt{1+\tan^2 \theta}.\sec^2 \theta. d \theta\]\[=\int\limits_{}{}\sec^3 \theta d \theta\]

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

    You can integrate this by parts. Set u = sec(theta) and dv = sec^2(theta). Then u = sec(theta)tan(theta) and v = tan(theta). Your integral becomes,\[\int\limits_{}{}\sec^3 \theta d \theta = \sec \theta \tan \theta -\int\limits_{}{}\sec \theta \tan^2 \theta d \theta\]

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

    \[=\sec \theta \tan \theta - \int\limits_{}{}\sec \theta (\sec^2 \theta -1) d \theta \]\[=\sec \theta \tan \theta - \int\limits_{}{}\sec^3 \theta d \theta+ \int\limits_{}{}\sec \theta d \theta\]

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

    Now \[\int\limits_{}{}\sec \theta d \theta = \ln|\sec \theta + \tan \theta|+c\] (I'm assuming you don't have to show that), so that your integral equation becomes,\[\int\limits_{}{}\sec^3 \theta d \theta = \sec \theta \tan \theta-\int\limits_{}{}\sec^3 \theta d \theta + \ln|\sec \theta + \tan \theta| + c\]

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

    Add the sec^3(theta) integral to both sides, \[2\int\limits_{}{}\sec^3 \theta d \theta =\sec \theta \tan \theta + \ln|\sec \theta + \tan \theta|+c\]

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

    and divide by 2 to get,\[\int\limits_{}{}\sec^3 \theta d \theta =\frac{1}{2}\sec \theta \tan \theta + \frac{1}{2}\ln|\sec \theta + \tan \theta|+c\]

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

    Now you have to use your relationship for x=tan(theta) and the relationship betwen sec(theta) and tan(theta) to find your result in terms of x.

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

    Now you have to use your relationship for x=tan(theta) and the relationship betwen sec(theta) and tan(theta) to find your result in terms of x.

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

    This site is causing me grief - won't post properly.

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

    i became a fan

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

    Awesome...cheers ;)

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

    You fine to finish it from here?

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

    yah ty

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

    No worries.

  25. 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.