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How to Anti-differentiate 1/sin(x) + cos(x)

  1. Apr 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Please find the anti-derivative of 1/sin(x) + cos(x) dx


    2. Relevant equations

    csc(x)dx = -ln[csc(x) + cot(x)] + C


    Thanks. Been at it for a while now.
     
  2. jcsd
  3. Apr 15, 2009 #2

    Cyosis

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    I am confused, you are given the primitive of the cosecant and you do not know how to integrate the cosecant+ the cosine? Or do you need to derive the primitive for the cosecant as well?
     
  4. Apr 15, 2009 #3

    Dick

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  5. Apr 15, 2009 #4
    Sorry it was my mistake. Its 1 over the sin and cos
     
  6. Apr 15, 2009 #5

    Mark44

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    Or in other words, the integrand is 1/(sin(x) + cos(x)).

    Parentheses are especially important when you're writing algebraic expressions on what is essentially a single line.

    By writing 1/sin(x) + cos(x), most people would correctly interpret this as
    [tex]\frac{1}{sin(x)} + cos(x)[/tex]
    even though that's not what you intended.
     
  7. Apr 15, 2009 #6
    yes i get it, but this has in no way helped me answer the question.
     
  8. Apr 15, 2009 #7

    Dick

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    Did you miss my response in post 3?
     
  9. Apr 15, 2009 #8

    Mark44

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    But it made it harder for people on this forum to understand exactly what you were asking. The way you wrote it confused Cyosis, although Dick was able to translate what you wrote into what the problem actually was. My post was aimed at getting you to realize the importance of writing your problems so that people can easily understand them.
     
  10. Apr 15, 2009 #9
    No, I read your post, but the questions asks to use the anti-derivative of csc(x) in order to solve the problem. :(

    Any ideas?
     
  11. Apr 15, 2009 #10
    yes i realized that, so i quickly posted that it is 1 over the sin and cos. So do you have any ideas on solving that with the anti-deriv of csc(x)?
     
  12. Apr 15, 2009 #11
    sin(x+y) = sin(x)cos(y) + cos(x)sin(y)

    So, putting y = pi/4 yields:

    sin(x)+cos(x) = sqrt(2) sin(x+pi/4)

    This means that you need to integrate 1/sin(x +pi/4) and you are already given the answer to that which you could also have derived very easily like this:

    1/sin(x) = [sin^2(x/2) + cos^2(x/2)]/[2 sin(x/2)cos(x/2)] =

    1/2 tan(x/2) + 1/2 cot(x/2) which is trivial to integrate.
     
  13. Apr 16, 2009 #12

    Mark44

    Staff: Mentor

    Inasmuch as you are studying calculus, it is to your benefit to learn how to express simple mathematical expressions clearly. In particular "1 over the sin and cos" is much less clear than "1 over the sum of the sine and cosine functions" or in symbols, 1/(sin(x) + cos(x)). The operation in the denominator is addition, not "and."
     
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