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Fairly complex trig integration

  1. Sep 29, 2009 #1
    1. The problem statement, all variables and given/known data
    [itex]\int dx/(cosx + cos3x)[/itex]

    2. Relevant equations

    3. The attempt at a solution
    Not sure if I'm doing this right.
    [itex]1/cosx + cos3x = 1/2cosx * cos2x = cosx/cosx * 2cosxcos2x = cosx/2cos^2xcosx = (1/2)*(cosx)/[(1-sin^2x)(1-2sin^2x)][/itex]

    then we let t = sinx and get [itex]\int dx/(cosx + cos3x) = 1/2 [1/(1-t^2)(1-2t^2)]dt[/itex]

    Any comments please..
  2. jcsd
  3. Sep 29, 2009 #2


    Staff: Mentor

    Please use more parentheses.

    How did you get from the first expression to the second?
    1/(cos(x) + cos(3x)) = 1/(2cos(x)cos(2x))
  4. Sep 29, 2009 #3
    it was a hint given by a TA.. I plugged in few numbers and the relationship seems to hold?
  5. Sep 30, 2009 #4


    Staff: Mentor

    It may be true, but you can't really prove that it is by merely plugging in a few numbers. In any case, it's not obvious, so you should put in the intervening steps.

    [tex]\frac{1}{cosx + cos3x}= \frac{1}{2cosx cos2x}= \frac{cos x}{cosx * 2cosxcos2x} = \frac{cos x}{2cos^2xcosx }= \frac{cosx}{2(1-sin^2x)(1-2sin^2x)}[/tex]

    Here's what you posted in the first post, using tex tags.
    Show how you get from the first expression to the second.
    From the 2nd to the 3rd, you apparently multiplied by cos x /cos x, which is fine.
    From the 3rd to the 4th is fine.
    From the 4th to the 5th, I think you have an error.

    Finally, in your integral, show the step where you make the substitution but before you actually integrate.
  6. Sep 30, 2009 #5
    Thank you for your reply.. I worked out a new solution, can u see if anything's wrong?
    \int \frac{dx} {cosx-cos3x} =
    \frac{1}{4} (-(\frac{1}{2}) cot(\frac{x}{2}) - ln(cos(\frac{x}{2}) - sin(\frac{x}{2})) + ln(cos(\frac{x}{2}) + sin(\frac{x}{2}) - (\frac{1}{2}) tan(\frac{x}{2})) + C
  7. Sep 30, 2009 #6


    Staff: Mentor

    Why don't you check it yourself? If your answer is correct, its derivative ought to be your integrand. In my previous post I asked you to show how you got from the first expression to the next in your string of equalities, but you haven't done so.
  8. Sep 30, 2009 #7
    using identities:
    cos(3x) = cos(2x+x) = cosx cos2x - sin^2x sinx
    cos(x) = cos(2x-x) = cosx cos2x + sin^2x sinx

    2nd term cancels out and we get 2cosxcos2x at the bottom. I got some help from my TA and solved it, thx for your help!
  9. Sep 30, 2009 #8


    Staff: Mentor

    I get
    cos(3x) = cos(2x+x) = cosx cos2x - sin2x sinx
    cos(x) = cos(2x-x) = cosx cos2x + sin2x sinx
    So cos(x) - cos(3x) = cosx cos2x + sin2x sinx - cosx cos2x + sin2x sinx = 2sin(2x) sin(x)

    That's not the same as what you show below...

  10. Sep 30, 2009 #9
    original question is actually cos(x) + cos(3x), not minus.. I made a mistake in post #5
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