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Math Help - Trig Integral

  1. #1
    Member VitaX's Avatar
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    Trig Integral

    \int \frac{sin(2x) - x}{x^2 + cos(2x)} dx

    No clue how to start this problem. Any help appreciated, I'll be looking through my notes to see if we've done something similar to this one in the mean time.
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  2. #2
    o_O
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    Try the substitution: u = x^2 + \cos (2x)

    See where that takes you
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  3. #3
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    Quote Originally Posted by o_O View Post
    Try the substitution: u = x^2 + \cos (2x)

    See where that takes you
    u=x^2+Cos(2x)

    du=2x - Sin(2x)dx

    But I cant see how it applies to the Integral because the coefficient 2 in front of the x in the derivative. Am I missing to see something here?
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  4. #4
    o_O
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    You made an error when you took the derivative of \cos (2x). Remember chain rule? A factor of 2 should then appear in your derivative which you can then factor out et voila. A standard integral.
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  5. #5
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    Quote Originally Posted by o_O View Post
    You made an error when you took the derivative of \cos (2x). Remember chain rule? A factor of 2 should then appear in your derivative which you can then factor out et voila. A standard integral.
    oh dang you're right I overlooked the 2. So the \frac{d}{dx} Cos(2x) = -Sin(2x)(2x)' \rightarrow -2Sin(2x)

    This correct?
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  6. #6
    o_O
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    Yup. So: du = \left(2x -2\sin (2x)\right)dx = -2 \left(\sin(2x) - x\right)dx

    which nicely appears in your numerator. Hence, a standard integral
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  7. #7
    Member VitaX's Avatar
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    Ok here's my work:


    u=x^2+Cos(2x)
    du=3x-2Sin(2x)dx \rightarrow du=-2(-x+Sin(2x))dx\rightarrow -\frac{1}{2}du=-x+Sin(2x)dx

    -\frac{1}{2}\int \frac{1}{u}du \rightarrow -\frac{1}{2}[ln|u|]

    -\frac{1}{2}ln|x^2+Cos(2x)|+C

    Hows it look?
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  8. #8
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    du={\color{red}3}x-2Sin(2x)dx \rightarrow du=-2(-x+Sin(2x))dx\rightarrow -\frac{1}{2}du=-x+Sin(2x)dx
    Just a little typo which you didn't carry through. Good job!
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