2
$\begingroup$

Integrate $$\int_0^{2\pi}\frac{1}{1-2p\cos A-p^2}\,\mathrm{d}A$$

Is there a standard method to solve integration problems of the above structure??

I tried substituting $p=\sin A$ but it was of no use.

  • 0
    with respect to $p$ or $A?$ – 2017-01-25
  • 0
    @labbhattacharjee with respect to A – 2017-01-25
  • 1
    Use https://en.wikipedia.org/wiki/Tangent_half-angle_substitution – 2017-01-25
  • 0
    @labbhattacharjee if you don't mind,can you be specific of which half angle substitution should be used here.thank you.. – 2017-01-25
  • 0
    Learn [MathJax](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) for posting here. – 2017-01-25
  • 0
    @StubbornAtom I'm sorry about the trouble.If you don't mind have a look at the image I've just included – 2017-01-25
  • 0
    $$1-p^2-2p\cos A=1-p^2-2p\cdot\dfrac{1-t^2}{1+t^2}=\dfrac{(1-p^2)(1+t^2)-2p(1-t^2)}{1+t^2}$$ $$=\dfrac{1-p^2-2p+t^2(1-p^2+2p)}{1+t^2}$$ where $t=\tan\dfrac A2$ – 2017-01-25
  • 0
    @labbhattacharjee thank you! – 2017-01-25
  • 0
    @labbhattacharjee If you don't mind,can you please tell me a way to integrate √(1+tan x)??? I tried with u=√(1+tan x) substitution, but it seems to be making the integral more complex.. – 2017-01-25
  • 0
    May I request you to post each problem in a different post along with your effort – 2017-01-25
  • 0
    @labbhattacharjee Sure... – 2017-01-25
  • 0
    Are you sure the denominator is $1-2\rho\cos(A)\color{red}{-}\rho^2$? A term $1-2\rho\cos(A)\color{red}{+}\rho^2$ would give you a $$\frac{1}{(1-\rho e^{iA})(1-\rho e^{-iA})}$$ that is simpler to integrate. – 2017-01-25
  • 0
    @JackD'Aurizio yeah I'm sure :) – 2017-01-25

1 Answers 1

3

Using half-angle substitution formulae to get $$I =-\int \frac {\sec^2 a/2}{p^2 \tan^2 a/2 -2p\tan^2 a/2 -\tan^2 a/2 + p^2 +2p-1} \mathrm {d}a $$ Substitute $u =\frac {\sqrt {p^2-2p-1}\tan a/2}{\sqrt {p^2+2p-1}} $, to get $$I =-\frac {2}{\sqrt {p^2+2p-1}\sqrt {p^2-2p-1}} \int \frac {1}{1+u^2} \mathrm {d}u $$

Hope you can take it from here.