I'm having some issues with the following integral
$$\int_{\frac{-\pi}{2}}^\frac{\pi}{2}\frac{\sec^2\theta}{1+\tan^2\theta \cos^2(2\alpha)}d\theta$$
My attempt is as follows, substitute $u=\tan\theta$(but this gives infinite bounds)
So $d\theta=\frac{1}{\sec^2\theta}du$, substituting both $\theta$ and $d\theta$ gives
$$\int_{\tan(\frac{-\pi}{2})}^{\tan(\frac{\pi}{2})}\frac{1}{1+u^2 \cos^2(2\alpha)}du$$
This time substituting $v=u\cos(2\alpha)$, $du=\frac{1}{\cos(2\alpha)}dv$, which gives
$$\int_{\tan(\frac{-\pi}{2})\cos(2\alpha)}^{\tan(\frac{\pi}{2})\cos(2\alpha)}\frac{1}{1+v^2}dv=\bigg{[} \arctan (v)\bigg{]}_{\tan(\frac{-\pi}{2})\cos(2\alpha)}^{\tan(\frac{\pi}{2})\cos(2\alpha)}$$
I don't think I've made any mistakes in my substitutions, but I'm still wondering how to get past the infinite bounds, since $\tan(\pi/2)=\infty$ and $\tan(-\pi/2)=-\infty$