# Math Help - Stuck on this integral

1. ## 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.

2. Originally Posted by dats13
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&#x5b;1&#x2f;&#x28;1 - A&#x2a;Sin&#x5b;x&#x5d;&#x29;&#x5d; - Wolfram|Alpha

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3. Is there simpler way of doing it? This way seems awfully complicated.

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

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

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

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