1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Trigonometry Substitution (Integral)

  1. Feb 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Making sure I got the right answer.


    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Feb 22, 2013 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Um, sec(theta)^2/sqrt(sec(theta)^2) isn't equal to 1.
     
  4. Feb 22, 2013 #3
    Hmm... I don't think so.

    This doesn't need a trig sub at all. Remember your basic algebra too!

    Hint: Look at the integral of the hyperbolic sine function, and see if there is something you can do to your function to make it a simple integral of that form.
     
  5. Feb 22, 2013 #4
    Correct, so its 1/4∫ secθ dθ

    which is ln | secθ + tanθ | + C

    1/4ln | (x+2/2) + (x/2) | is the answer right?
     
  6. Feb 22, 2013 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I think it's pretty likely that KDeep hasn't done hyperbolic trig functions yet, though that does make it easier. You need some kind of substitution. Ordinary trig will work fine. You just have to do it right.
     
  7. Feb 22, 2013 #6
    I have never heard about hyperbolic sine function at all, I will definitely look into that. My Professor wants us to do trig substitutions since it is in the section.
     
  8. Feb 22, 2013 #7
    When you take the derivative of your solution, do you get the function you wished to integrate?
     
  9. Feb 22, 2013 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Getting close. Try the sec(arctan(x/2)) part again.
     
  10. Feb 22, 2013 #9
    I am confident that the derivative of ln| secθ + tanθ | is secθ.
     
  11. Feb 22, 2013 #10
    You are finding the antiderivative of 1/sqrt(16+4x^2), so your solution must yield 1/sqrt(16+4x^2) (or equivalent) when it is differentiated.

    I misspoke anyway - I meant inverse hyperbolic sine, but yes, you're right.
     
  12. Feb 22, 2013 #11
     
  13. Feb 22, 2013 #12
    It doesn't matter. Given those replacements, replace them back when you have your solution.
     
  14. Feb 22, 2013 #13

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You are almost there. You got the sec(arctan(x/2)) part wrong and you need to correct a simple numerical mistake. That's all.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Trigonometry Substitution (Integral)
Loading...