Question

On computing the following closed line integral.

Answer 1

Given a path, compute the integral:
\[
\oint_\gamma \frac{xdy-ydx}{x^2+y^2}
\]Where the path is:
\[
\langle x,y\rangle=\langle 1+2\cos t, 2\sin t\rangle, 0\le t\le 2\pi
\]

Answer 2

It's pretty clear that the integral is path-independent with the exclusion of a residue if \(\gamma\)'s interior contains the origin. So the integral evaluates to \(2\pi\). Sadly, though, I reach this step in integration and cannot proceed further:
\[
\int_{[0,2\pi]}dt\;\frac{2\cos t+4}{4\cos t+5}
\]
Anyone have any ingenious substitutions for this?

Answer 3

@satellite73 @ganeshie8

Answer 4

check your numerator once

Answer 5

\[\large \langle x,y\rangle=\langle 1+2\cos t, 2\sin t\rangle\]

\[\large \langle dx,dy\rangle=\langle -2\sin t, 2\cos t\rangle\]

Answer 6

Oh, no, the integral is correct; I just don't by what substitution to solve it. Not in a direct way, at least, with the given path.

Answer 7

\[\oint_\gamma \frac{xdy-ydx}{x^2+y^2} = \oint_\gamma \frac{(1+2\cos t)(2\cos t) - 2\sin t(-2 \sin t)}{(1+2\cos t)^2 + (2\sin t)^2} dt\]

Answer 8

Yes... now if we simplify that down, and we receive the above integral.

Answer 9

ahh ok, i made a mistake earlier... nvm

Answer 10

\[\large \int_{0}^{2\pi}\;\frac{2\cos t+4}{4\cos t+5} dt\]

Answer 11

Point is, proving the above is very simple if we just prove that, for some arbitrary curve bounding the origin, the integral is equivalent to a circle around the origin, which is, of course equal to \(2\pi\). I want to evaluate this directly, though, and it is proving to be much more difficult.

Answer 12

try the tangent half-angle substitution

Answer 13

or "Weierstrass substitution"

Answer 14

https://www.wolframalpha.com/input/?i=integrate+%282cos%28x%29+%2B4%29%2F%284cos%28x%29%2B5%29dx

Answer 15

@Zarkon trying it.

@aum Yes, I know what it evaluates to. Read the above.

Answer 16

\[
2\cos(t)+4=\frac{1}{2}\left(4\cos(t)+5\right)+\frac{3}{2}?
\]

Answer 17

Already tried that, but thank you @thomas5267 .

Answer 18

that simplifies nicely actually :

\[\large \int_{0}^{2\pi}\;\frac{2\cos t+4}{4\cos t+5} dt\]

\[\large \int_{0}^{2\pi}\;\frac{\frac{1}{2}\left(4\cos(t)+5\right)+\frac{3}{2}}{4\cos t+5} dt\]

\[\large \frac{1}{2} \int_{0}^{2\pi} dt + \frac{3}{2}\int_{0}^{2\pi}\;\frac{1}{4\cos t+5} dt\]

Answer 19

\[\int_{0}^{2\pi}\;\frac{1}{4\cos t+5} dt \]

\[\int_{0}^{2\pi}\;\frac{1}{4(\cos^2(t/2) - \sin^2(t/2))+5(\cos^2(t/2) + \sin^2(t/2)} dt \]

\[\int_{0}^{2\pi}\;\frac{1}{9\cos^2(t/2 )+ \sin^2(t/2)} dt \]

Answer 20

\[\frac{1}{9}\int_{0}^{2\pi}\;\frac{\sec^2(t/2)}{1+ \frac{1}{9}\tan^2(t/2)} dt \]

Answer 21

\(u = 3\tan (t/2)\)

\(\cdots\)

Answer 22

@Zarkon , you are a genius. It takes some awful amount of simplification, but it works. Also, @ganeshie8 and @thomas5267 , your solution works quite nicely, too. Fantastic, thanks!

Answer 23

you may like to go through this MSE link
http://math.stackexchange.com/questions/263397/how-to-evaluate-int-02-pi-fracd-thetaab-cos-theta

Answer 24

Just did, the application of the Cauchy integral is quite nice. It's a nicer result than I expected, I must say.

Thanks, again, mate.