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Math Help - Stuck on this integral

  1. #1
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    Stuck on this integral

    I am stuck on this integral and not sure how to go about it.





    where A is a constant. I have tried multiply top and bottom by , but I unsure of how to proceed from there as I cannot use a trig identity. Any advice would be appreciated.
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  2. #2
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    Quote Originally Posted by dats13 View Post
    I am stuck on this integral and not sure how to go about it.





    where A is a constant. I have tried multiply top and bottom by , but I unsure of how to proceed from there as I cannot use a trig identity. Any advice would be appreciated.
    integral[1/(1 - A*Sin[x])] - Wolfram|Alpha

    Click on Show Steps
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  3. #3
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    Is there simpler way of doing it? This way seems awfully complicated.
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  4. #4
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    Hello, dats13!

    Sorry, there is no simple way . . .

    Before you were assigned this integral,
    . . you should have been taught a particular substitution.

    The process is long and intricate, but it's your only chance to integrate it.
    (Well, if you're using only your brain, that is.)


    \int\frac{d\theta}{1 - A\cos\theta}
    I'll give you the entire derivation of this bizarre substitution.


    Let: . z \:=\:\tan\tfrac{\theta}{2} \quad\Rightarrow\quad \tfrac{\theta}{2} \:=\:\arctan z \quad\Rightarrow\quad \theta \:=\:2\arctan z

    . . Then: . d\theta \:=\:\frac{2\,dx}{1+z^2}\;\;{\color{blue}[1]}

    We have: . \tan\tfrac{\theta}{2} \:=\:\frac{z}{1} \:=\:\frac{opp}{adj}

    So \tfrac{\theta}{2} is in a right triangle with: opp = z,\;adj = 1
    And Pythagorus says: . hyp = \sqrt{1+z^2}

    Hence: . \sin\tfrac{\theta}{2} \:=\:\frac{z}{\sqrt{1+z^2}},\;\;\cos\tfrac{\theta}  {2} \:=\:\frac{1}{\sqrt{1+z^2}}

    . . Then: . \begin{array}{ccccccccc}\sin\theta &=&2\sin\frac{\theta}{2}\cos\frac{\theta}{2} &=& 2\cdot\dfrac{z}{\sqrt{1+z^2}}\cdot\dfrac{1}{\sqrt{  1+z^2}} &=&\dfrac{2z}{1+z^2} & {\color{blue}[2]} \\ \\[-3mm]<br />
\cos\theta &=& \cos^2\!\frac{\theta}{2} - \sin^2\!\frac{\theta}{2} &=& \left(\dfrac{1}{\sqrt{1+z^2}}\right)^2 - \left(\dfrac{z}{\sqrt{1+z^2}}\right)^2 &=& \dfrac{1-z^2}{1+z^2} & {\color{blue}[3]} \end{array}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Substitute [1] and [3] into the integral:

    . . \int\frac{\frac{2\,dx}{1+z^2}}{1 - A\frac{2z}{1+z^2}} \;=\; 2\int\frac{dz}{z^2-2Az + 1}

    The denominator is: . z^2 - 2Az \:{\color{red}+ \:A^2} + 1 \:{\color{red}-\: A^2} \;=\;(z-A)^2 + (1-A^2)


    Hence, the integral is: . 2\int \frac{dx}{(z-A)^2 + \left(\sqrt{1-A^2}\right)^2} . . . arctangent form


    So we have: . \frac{2}{\sqrt{1-A^2}}\arctan\left(\frac{z-A}{\sqrt{1-A^2}}\right)+C


    Back-substitute: . \frac{2}{\sqrt{1-A^2}}\arctan\left(\frac{\tan\frac{\theta}{2} - A}{\sqrt{1-A^2}}\right) + C

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  5. #5
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    Thanks for the help, although I wasn't expecting anything like that.
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