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.
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.
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.)
I'll give you the entire derivation of this bizarre substitution.$\displaystyle \int\frac{d\theta}{1 - A\cos\theta}$
Let: .$\displaystyle z \:=\:\tan\tfrac{\theta}{2} \quad\Rightarrow\quad \tfrac{\theta}{2} \:=\:\arctan z \quad\Rightarrow\quad \theta \:=\:2\arctan z$
. . Then: .$\displaystyle d\theta \:=\:\frac{2\,dx}{1+z^2}\;\;{\color{blue}[1]}$
We have: .$\displaystyle \tan\tfrac{\theta}{2} \:=\:\frac{z}{1} \:=\:\frac{opp}{adj}$
So $\displaystyle \tfrac{\theta}{2}$ is in a right triangle with: $\displaystyle opp = z,\;adj = 1$
And Pythagorus says: .$\displaystyle hyp = \sqrt{1+z^2}$
Hence: .$\displaystyle \sin\tfrac{\theta}{2} \:=\:\frac{z}{\sqrt{1+z^2}},\;\;\cos\tfrac{\theta} {2} \:=\:\frac{1}{\sqrt{1+z^2}}$
. . Then: . $\displaystyle \begin{array}{ccccccccc}\sin\theta &=&2\sin\frac{\theta}{2}\cos\frac{\theta}{2} &=& 2\cdot\dfrac{z}{\sqrt{1+z^2}}\cdot\dfrac{1}{\sqrt{ 1+z^2}} &=&\dfrac{2z}{1+z^2} & {\color{blue}[2]} \\ \\[-3mm]
\cos\theta &=& \cos^2\!\frac{\theta}{2} - \sin^2\!\frac{\theta}{2} &=& \left(\dfrac{1}{\sqrt{1+z^2}}\right)^2 - \left(\dfrac{z}{\sqrt{1+z^2}}\right)^2 &=& \dfrac{1-z^2}{1+z^2} & {\color{blue}[3]} \end{array}$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Substitute [1] and [3] into the integral:
. . $\displaystyle \int\frac{\frac{2\,dx}{1+z^2}}{1 - A\frac{2z}{1+z^2}} \;=\; 2\int\frac{dz}{z^2-2Az + 1}$
The denominator is: .$\displaystyle z^2 - 2Az \:{\color{red}+ \:A^2} + 1 \:{\color{red}-\: A^2} \;=\;(z-A)^2 + (1-A^2) $
Hence, the integral is: .$\displaystyle 2\int \frac{dx}{(z-A)^2 + \left(\sqrt{1-A^2}\right)^2} $ . . . arctangent form
So we have: .$\displaystyle \frac{2}{\sqrt{1-A^2}}\arctan\left(\frac{z-A}{\sqrt{1-A^2}}\right)+C $
Back-substitute: .$\displaystyle \frac{2}{\sqrt{1-A^2}}\arctan\left(\frac{\tan\frac{\theta}{2} - A}{\sqrt{1-A^2}}\right) + C$