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Evaluate integral

  1. Feb 6, 2009 #1
    1. The problem statement, all variables and given/known data

    between pi/2 and pi/6[tex]\int[/tex] (cos x) / (7 + sin x)

    move 1/7 to the out side

    1/7 [tex]\int[/tex] cos x / sin x

    u= sin x
    du= cos x

    so i get
    1/7 ln sinx + C

    then plug pi/2 minus pi/6???
  2. jcsd
  3. Feb 6, 2009 #2


    Staff: Mentor

    You can't do that.

    Is [tex]\frac{1}{7 + 3} = 1/7 * \frac{1}{3}[/tex]

    Instead, you might try multiplying by (7 - sin(x))/(7 - sin(x)).
  4. Feb 6, 2009 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    If you are taking calculus, your algebra should be better than that! cos(x)/(7+ sin(x)) is NOT (1/7) cos(x)/sin(x).

    Instead use the substitution u= 7+ sin x.

    Mark got in just ahead of me but I think my suggestion for integrating it is easier!
    Last edited: Feb 6, 2009
  5. Feb 6, 2009 #4
    k got it
    i have taken any math for 2 year and i am now taking cal 2 so yea... but i solved it thanks
  6. Feb 6, 2009 #5


    Staff: Mentor

    Yes, I agree. It just goes to show that "haste makes waste."
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