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Integrals + Trig

  1. Jan 24, 2008 #1
    I was reviewing my first calc class stuff before starting this second one and came across a problem that i cant seem to get, its the integral of sin(2x)/23+cos(x)^2 dx, i know most of the rules and thought i had it but the question asks to put in all trig functions in terms of cos which I cant seem to figure out how to do. Been awhile since I've done this stuff so sorry if its real easy and im just missing something simple. thanks in advance.
  2. jcsd
  3. Jan 24, 2008 #2


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    How about let [itex]u=23+cos^2x[/itex]
  4. Jan 24, 2008 #3
    What does the double angle identity [tex]\sin{2x}[/tex] simplify to?
  5. Jan 25, 2008 #4


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    Is this a "differential equations" problem?
  6. Jan 25, 2008 #5
    no i dont believe so, its an integral problems that i jus dont understand thought this would be the best place to put it.
  7. Jan 26, 2008 #6
    What exactly is the question? Is it:

    [tex]A=\int \frac{sin(2x)}{23+cos(x^2)}dx[/tex]


    [tex]B=\int \left(\frac{sin(2x)}{23}+cos(x^2) \right)dx[/tex]


    [tex]C=\int \frac{sin(2x)}{23+cos^2(x)}dx[/tex]


    [tex]D=\int \left(\frac{sin(2x)}{23}+cos^2(x) \right)dx[/tex]

    It is unclear what you mean.
  8. Jan 26, 2008 #7

    Gib Z

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    Almost certainly C. Though I get your point, careless notation is annoying.
  9. Jan 26, 2008 #8
    it is the C one u posted IDK how to make it clearer writing all these functions n stuff out with a keyboard
  10. Jan 26, 2008 #9
    This is a link where you can find the basics:


    If you click on a formula, the latex code pops up. No worries about the typing, you'll learn it. So the third one is the one you need to tackle. What have you got so far?
  11. Jan 26, 2008 #10


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    In that case consider re-writing the denominator;

    [tex]23+cos^2(x) = 23 + \frac{1}{2}\left(1 + \cos(2x)\right) = \frac{1}{2}\left(47+\cos(2x)\right)[/tex]

    Now take the derivative and compare with the numerator.
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