I'm trying to calculate this integral
$\int_0^{2π}{1 \over{(Acos^2t+Bsin^2t)^2}}dt$, where $A,B>0$ using differetntial forms.
I have calculated $\int_0^{2π}{1 \over{(Acos^2t+Bsin^2t)}}dt={2π\over {\sqrt{AB}}}$ (I hope it's correct) and tried using it but that squared part is messing with me and I've also tried showing the form $-{y\over{(x^2+y^2)^2}}dx+{x\over{(x^2+y^2)^2}}dy$ is closed, but it turned out it isn't. So I'm stuck.
Can you give me any hints but please not solutions?
Firstly you notice that $\cos^2t=\frac{1+\cos2t}{2}$ and $\sin^2t=\frac{1-\sin2t}{2}$, so your denorminator becomes $(A+B\cos2t)^2$ for some $A,B$. Then using changing of variable your denominator becomes $(A+B\cos t)^2$ and you calculate the integral over $[0,\pi]$ and times factor $1/2$. Now the denorminator becomes $(A^2+B^2)\cos^2(t+\theta)$ for some $\theta$ depending on $A$ and $B$. Finally using $(\tan t)'=\frac{1}{\cos^2 t}$ you have the antiderivative and get your result.