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Math Help - Integration

  1. #1
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    Question Integration

    What is the integral of: (4sin(x))/(2+cos(x)) and please explain the method. Thank you.
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  2. #2
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    Hello, charlie8529!

    \int \frac{4\sin x}{2+\cos x}\,dx

    We have: . 4 \int\frac{1}{2+\cos x}\,(\sin x\,dx)


    Let: u \:=\:2 + \cos x\quad\Rightarrow\quad du \:=\:-\sin x\,dx \quad\Rightarrow\quad \sin x\,dx \:=\:-du

    Substitute: . 4\int\frac{1}{u}\,(-du) \;=\;-4\int\frac{du}{u}


    Got it?

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  3. #3
    Senior Member vincisonfire's Avatar
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    You want to compute the antiderivative of
     \frac{4sin(x)}{2+cos(x)}
    That is  \int \frac{4sin(x)}{2+cos(x)}dx .
    Let  u = 2 + cos(x)
    Then you can differentiate both side to find that
     du = -sin(x)dx
    We notice that in the integral above we have  4 sin(x)dx at the denominator.
    This corresponds to  -4du = -4(-sin(x))dx = 4 sin(x) dx
    We can now substitute in the integral to simplify the notation.
     \int \frac{4sin(x)}{2+cos(x)}dx = \int \frac{-4du}{u}dx = -4 \int \frac{du}{u}dx
    This should now remember you something.
     -4 \int \frac{du}{u}dx = -4 ln|u| = -4 ln|2+cos(x)| because we substitute math] u = 2 + cos(x) [/tex].
    You may use logarithm properties to express your answer as
     ln|\frac{1}{(2+cos(x))^4}|+C where C is a constant number. Don't forget it. I use to and teachers will take off mark any time for this little mistake.
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  4. #4
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    Thanks for the help guys.
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