How to Solve an Integral Using t-Substitution?

[noospace]

Homework Statement

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

The Attempt at a Solution

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

<br /> t^2 + 1 = \sec^2\alpha \implies \cos^2\alpha = \frac{1}{t^2 +1}<br />
<br /> \sin^2\alpha = t^2\cos^2\alpha = \frac{t^2}{t^2+1}<br />
<br /> dt = \sec^2\alpha d\alpha \implies d\alpha = \frac{dt}{1 + t^2}<br />

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)}

<br /> 1 = C_1(A+ (1+A)t^2) + C_2 (1+t^2) \implies C_1 = -1,C_2 = -\frac{1+A}{A}<br /> [\latex]<br /> <br /> I = \int_{0}^\infty \frac{-1}{1+t^2}- \frac{(1+A)/A}{A+(1+A)t^2}<br /> I = - \frac{\pi}{2}- (1+A)/A\int_{0}^\infty\frac{dt}{A+(1+A)t^2}<br /> 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<br /> <br /> I = - \frac{\pi}{2}- \frac{(1+A)/A}{\sqrt{1+A}}\frac{1}{\sqrt{A}}\frac{\pi}{2}<br /> <br /> Thanks.

 

[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}.