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I'm having a trouble with this integral expression:

$$\int_0^{2\pi} \frac{d\theta}{A+B \cos\theta}$$

I've done this substitution: $t= \tan(\theta/2)$

and get: $\displaystyle \cos\theta= \frac{1-t^2}{1+t^2}$ and $\displaystyle d\theta=\frac{2}{1+t^2}dt$ where $\displaystyle \cos^2\theta/2=\frac{1}{1+t^2}$

then the integral becomes:

$$\int\frac{2 \, dt}{(A-B)t^2+(A+B)}= \sqrt\frac{A+B}{A-B} \arctan \left(\left(\sqrt\frac{A+B}{A-B} \right) t\right)$$

However, I'm not sure about the new limits since $\tan$ has period $\pi$ so what I have to do at this point to decide the new limits? And of course if you find some mistake in what I've done before, please let me know!

  • 4
    Hint: $ \displaystyle \int_0^{2\pi} \frac{d\theta}{A+B\cos\theta} = 2 \displaystyle \int_0^{\pi} \frac{d\theta}{A+B\cos\theta}.$2012-12-21
  • 2
    Are you dealing with explicit $A$ and $B$? Sometimes the integral will not even converge.2012-12-21
  • 1
    By the way, you don't have to use Weierstrass substitution that explicitly since we can write: $$ \begin{aligned}I & = \frac{1}{a+b\cos{x}}\\& =\frac{1}{a\left(\sin^2 \frac{1}{2}x+\cos^2\frac{1}{2}x\right)+b\left(\cos^2 {\frac{1}{2}x}-\sin^2\frac{1}{2}x\right)} \\& =\frac{1}{(a-b)\sin^2\frac{1}{2}x+(a+b)\cos^2\frac{1}{2}x} \\& =\frac{\sec^2{\frac{1}{2}x}}{(a+b)+(a-b)\tan^2\frac{1}{2}x}. \end{aligned}$$2012-12-21

4 Answers 4