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Integration problem in trigonometric functions

  1. Nov 25, 2013 #1
    1. The problem statement, all variables and given/known data

    please help me with this integration problem?

    ∫(1/sinx+ cosx) dx

    2. Relevant equations

    i dont know any proper substitution in this question,maybe there are none

    3. The attempt at a solution

    i tried rationalizing and it has got me this far.. ∫(sinx-cosx)/[(sinx)^2 - (cosx^2)] dx...please help me
     
  2. jcsd
  3. Nov 25, 2013 #2
    better to integrate the terms separately. The cos(x) is no problem. The other term is csc(x). Here's how you go about it:

    ##\int## csc(x) dx =## \int## csc(x) ## \cdot \frac{csc x + cot x}{ csc x + cot x}##dx
    use the substitution:

    u = csc x + cot x

    You say that's not obvious? Right -- it is a trick. There are thousands of them, worked out over many years by lots of smart mathematicians. This one is more commonplace than some.
     
  4. Nov 25, 2013 #3

    Mark44

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    brmath, I don't see that this will help.
    I believe that this is the integral in question:
    $$ \int \frac{dx}{sin(x) + cos(x)}$$

    kashan, brmath interpreted what you wrote (1/sin(x)) + cos(x), which from your work, wasn't what you intended. To prevent this kind of confusion, place parentheses around the entire denominator, like this: 1/(sin(x) + cos(x))
     
    Last edited: Nov 25, 2013
  5. Nov 25, 2013 #4
  6. Nov 25, 2013 #5
    I did so interpret it. Sorry.
     
  7. Nov 25, 2013 #6

    LCKurtz

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    No need to apologize. You worked what was actually posted. kashan123999 caused the confusion with his improperly written post.
     
  8. Nov 25, 2013 #7

    Mark44

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    Yes, no need for apologies, brmath. It takes some practice here to understand what neophytes mean, as opposed to what they actually write. The only way I could figure out what the OP actually meant was from the work he showed.
     
  9. Nov 25, 2013 #8

    lurflurf

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    hint
    $$\frac{1}{\sin(x)+\cos(x)}=\frac{\sqrt{2}}{2}\csc (x+\pi/4)
    $$
     
  10. Nov 25, 2013 #9

    Mark44

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    As one of my calculus instructors was fond of saying, "Intuitively obvious to the most casual observer."
     
  11. Nov 25, 2013 #10
    Well then, this makes it quite simple doesn't it...
     
  12. Nov 25, 2013 #11
    His idea of casual observer being not exactly applicable to people like me.
     
  13. Nov 25, 2013 #12

    LCKurtz

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    Has anyone but me noticed the OP never returned to this thread?
     
  14. Nov 26, 2013 #13

    Mark44

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    So let's hold with any more help until we hear back from the OP.
    I assure you, he was speaking ironically...
     
  15. Nov 26, 2013 #14
    of course he was speaking ironically -- so was I.
     
  16. Dec 1, 2013 #15
    I am extremely sorry for that misplacement of parenthesis that caused the whole confusion,very very sorry to all of you...it is yes it is actually ∫[1/(sinx + cosx)] dx so then,i have corrected it,can anyone tell me some easy or intuitive substitution,cause our teacher substituted 1 for some angle of sin i guess...and i am not able to grasp that...
     
  17. Dec 1, 2013 #16

    LCKurtz

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    Look at post #4 and post #8. Two different ways to attack the problem. You might have additional questions after looking at those posts.
     
  18. Dec 1, 2013 #17

    lurflurf

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    From your step
    (sinx-cosx)/[(sinx)^2 - (cosx^2)]
    =sinx/[(sinx)^2 - (cosx^2)]-cosx/[(sinx)^2 - (cosx^2)]
    you can substitute u=sin(x)in one and u=cos(x) in the other and use
    1=sin^2(x)+cos^2(x)
    and
    the integration formula for 1/(1-a^2 x^2)
    or use my identity above
    $$\frac{1}{\sin(x)+\cos(x)}=\frac{\sqrt{2}}{2}\csc (x+\pi/4)
    $$
    along with the integration formula for csc(x)
     
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