Question

Solve using contour integrals.
integral [0,pi] of (d(theta))/(A-Bcos(theta))

Answer 1

\[\int\limits_{0}^{\pi}\frac{ 1 }{ (A-Bcos(\theta)) }\]

Answer 2

If you have another method, then that is fine also.

Answer 3

is this simpler to line integrals?

Answer 4

\[\int_C~ f(x,y)~ds\]
\[\int\int~ f(x,y)~\sqrt{(x')^2+(y')^2}~dy~dx\]

Answer 5

well, the page im looking at uses dt instead of dydx a spose since its a parametric

Answer 6

Wait, if it was parametric, wouldn't the problem get more cmplicated?

Answer 7

lol, at the moment i dont know

Answer 8

hmmm, i tried taking the antiderivative, but that method isn't pretty here...

Answer 9

http://www1.mat.uniroma1.it/~isopi/didattica/vc/Trapper.pdf

this page may be useful, trying to review it

Answer 10

I wouldn't take an anti derivative. Or perhaps on my paper, not to put it in here. I wish I could help. When I actually learn some math, I'll be hopefully able to help by then.

Answer 11

hmmm, let me try it! can we both do it, so we can check each other's answer?

Answer 12

so first of, would I make the substitution costheta = ((1/2) (z + 1/2))

Answer 13

I think the sub tan(x/2)=theta might work

Answer 14

Um, i can't simply it enough, any more help?

Answer 15

\[\tan(\frac{x}{2})=\theta \]

\[\sin(x)=2 \sin(\frac{x}{2})\cos(\frac{x}{2})=2\frac{\theta}{\sqrt{\theta ^2 +1 }}\frac{1}{\sqrt{\theta ^2+1}}=\frac{2 \theta }{ \theta ^2+1} \\ \cos(x)=\cos^2(\frac{x}{2})-\sin^2(\frac{x}{2})=\frac{1}{\theta ^2 +1 } -\frac{\theta ^2}{\theta ^2 +1 } =\frac{1- \theta ^2 }{\theta ^2 +1} \\ \frac{1}{2} \sec^2(\frac{x}{2})dx=d \theta \\ dx =2 \cos^2(\frac{x}{2}) d \theta \\ dx =2 \frac{1}{\theta ^2 +1 } d \theta \\ dx=\frac{2 }{ \theta ^2 +1 } d \theta \]
\[\int\limits_{}^{}\frac{1}{A-B \cos(x)} d x =\int\limits_{}^{}\frac{1}{A-B \cdot \frac{1-\theta ^2}{1+\theta ^2}} \frac{2}{\theta ^2 +1} d \theta \]