1. Oct 19, 2007

noospace

1. The problem statement, all variables and given/known data

$\int_{0}^{2\pi} \frac{\cos^2\alpha}{A + \sin^2\alpha}d\alpha$

3. The attempt at a solution

I believe when the trig functions all appear squared one may use the substitution $t =\tan \alpha$. Then

$t^2 + 1 = \sec^2\alpha \implies \cos^2\alpha = \frac{1}{t^2 +1}$
$\sin^2\alpha = t^2\cos^2\alpha = \frac{t^2}{t^2+1}$
$dt = \sec^2\alpha d\alpha \implies d\alpha = \frac{dt}{1 + t^2}$

Now use

$\int_{0}^{2\pi} = \int_0^\pi + \int_\pi^{2\pi}$.

$\int_{0}^{\pi} \frac{\cos^2\alpha}{A + \sin^2\alpha}d\alpha = \int_{0}^\infty\frac{dt}{A(1+t^2)^2 +t^2(1+t^2)}$
$= \int_{0}^\infty\frac{dt}{(1+t^2)(A+t^2+At^2)}$

$1 = C_1(A+ (1+A)t^2) + C_2 (1+t^2) \implies C_1 = -1,C_2 = -\frac{1+A}{A} [\latex] [itex]I = \int_{0}^\infty \frac{-1}{1+t^2}- \frac{(1+A)/A}{A+(1+A)t^2}$
$I = - \frac{\pi}{2}- (1+A)/A\int_{0}^\infty\frac{dt}{A+(1+A)t^2}$
$I = - \frac{\pi}{2}- \frac{(1+A)/A}{\sqrt{1+A}}\frac{1}{\sqrt{A}}\left.\tan^{-1}\frac{t}{\sqrt{A}}\right|_{0}^\inft$

$I = - \frac{\pi}{2}- \frac{(1+A)/A}{\sqrt{1+A}}\frac{1}{\sqrt{A}}\frac{\pi}{2}$

Thanks.

Last edited: Oct 19, 2007
2. Oct 20, 2007

dynamicsolo

I'm with you up to the point where you derive the constants for the integration by partial fractions. (And I dropped the quotation because there's a hidden character that's really messing everything up here...)

For the $$t^{2}$$ terms, you have

$$C_{1}(1+A) + C_{2} = 0$$

and for the constant terms,

$$C_{1}A + C_{2} = 1$$.

So $$C_{2} = 1 - AC_{1}$$,

giving $$C_{1} + AC_{1} + ( 1 - AC_{1} ) = 0$$.

Sure, you get $$C_{1} = -1$$, but wouldn't that lead to just

$$C_{2} = 1 - A(-1) = 1 + A$$?

As another suggestion: upon plotting this function, I find that the area from x = 0 to x = $$2\pi$$ is four times the area from x = 0 to x = $$\frac{\pi}{2}$$.

Last edited: Oct 20, 2007