Help finding a method for integrating this trigonometry function I was going though a given proof in my book and part of it requires solving
$$\frac{ad-bc}{2 \pi} \int^{2\pi}_0 \frac{d \theta}{(a \cos\theta+b\sin\theta)^2+(c \cos\theta+d\sin\theta)^2} $$
Which it said using "standard techniques" would result in = sign($ad-bc$)
Maybe my integration techniques are a little rusty, but I was wondering how I go about this. I figure that the integration part somehow turns into something along the lines of $\dfrac{2\pi}{|ad-bc|}$ somehow, but I can't see about how to get there.
 A: Here is a geometric solution (which is not what you might want): Let $E$ be the image of the unit circle under the linear map with the matrix $M=\begin{pmatrix}a&b\\c&d\end{pmatrix}^{-1}$.
It is an ellipse, which can be parametrized as $(r(\theta)\cos\theta,r(\theta)\sin\theta)^T$ (for the appropriate function $r(\theta)$). Then $\int_0^{2\pi}r(\theta)^2\,d\theta$ is twice its area, i.e. $2\pi|\det M|$. 
To determine $r(\theta)$: the image of $(r(\theta)\cos\theta,r(\theta)\sin\theta)^T$ by $M^{-1}$ is on the unit circle, from which we get that $r(\theta)$ is the inverse of length of 
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}\cos\theta\\\sin\theta\end{pmatrix}$$
i.e. $\int_0^{2\pi}r(\theta)^2\,d\theta$ is precisely your integral.
A: Here is one approach that you may try: Remove brackets in the denominator. You will get a $2absint$ as well as a $2cdsint$ which can be converted into $(ab+cd)sin2t$ Further you will get a bunch of $sin^2t$ and $cos^2t$ terms that can all be converted into $cos2t$ terms by identities. Now your denominator is going to look like $p + qsin2t+rcos2t$  At this point you can use the tangent half angle substitution (Weierstrass substitution) to convert your integral into a rational function integral. Your denominator will be of quadratic nature and the numerator is some constant. Now it becomes straight forward. Mind you that you have to break up the integral into seperate integrals since there are values for which $tan(x/2)$ (your weierstrass sub) does not exist.
