Compute $I = \int_0^{2\pi} \frac{ac-b^2}{[a \sin(t)+ b \cos(t)]^2+[b \cos(t)+ c \sin(t)]^2}dt$ How do you compute
$$I = \int_0^{2\pi} \frac{ac-b^2}{[a \sin(t)+ b \cos(t)]^2+[b \cos(t)+ c \sin(t)]^2}dt = 2 \pi \ \mathrm{sign}(ac-b^2) $$
which comes up when computing the Poincare index? Wolfram does not work out at all or give the steps :(
It comes from Courant/John volume 2 page 357-358
http://i.stack.imgur.com/5kexM.png
 A: I will show you an overkill, but a useful one (at least, IMHO). If the quadratic form
$$ q(x,y) = A x^2 + 2B xy + C y^2 $$
associated with the matrix
$$ M_q=\begin{pmatrix} A & B \\ B & C\end{pmatrix}$$
is positive definite, i.e. (by Sylvester's criterion) $A>0$ and $AC-B^2>0$, then the integral
$$ I_q = \iint_{\mathbb{R}^2}\exp\left(-q(x,y)\right)\,dx\,dy $$
is convergent and its value equals $\frac{\pi}{\sqrt{\det M_q}}=\frac{\pi}{\sqrt{AC-B^2}}$. The proof relies on the spectral theorem, Fubini's theorem and the fact that the determinant of $M_q$ is the product of its eigenvalues. What happens if we perform a change of variables, for instance if we switch to polar coordinates? We get:
$$ I_q = \int_{0}^{2\pi}\int_{0}^{+\infty}\rho\exp\left(-\rho^2 q(\cos\theta,\sin\theta)\right)\,d\rho\,d\theta=\frac{1}{2}\int_{0}^{2\pi}\frac{d\theta}{q(\cos\theta,\sin\theta)}.$$
In your case, the coefficients of the quadratic form $q$ are:
$$ A=2b^2,\qquad B=ba+bc,\qquad C=a^2+c^2.$$
Can you guess now what the value of your integral has to be? I got:

$$ \int_{0}^{2\pi}\frac{dt}{(a\sin t+b\cos t)^2+(b\cos t+c\sin t)^2}=\color{red}{\frac{2\pi}{\left|b\right|\cdot\left|a-c\right|}}$$
  under the assumptions $b\neq 0$ and $a\neq c$.

