I am trying to do the following integral:
$$I=\int^{\pi}_{-\pi}\frac{1}{a\cos^2{2x}+1}\,dx$$
I am proceeding in the following manner: $$ \begin{align} I &=\int^{\pi}_{-\pi}\frac{1}{a\cos^2{2x}+1}\,dx\\ &=\int^{\pi}_{-\pi}\frac{\sec^2{2x}}{a+\sec^2{2x}}\,dx\\ &=\int^{\pi}_{-\pi}\frac{\sec^2{2x}}{a+1+\tan^2{2x}}\,dx\\ &=\int^{\pi}_{-\pi}\frac{\sec^2{2x}}{({\sqrt{a+1}})^2+\tan^2{2x}}\,dx\\ &=\frac{1}{2}\int^{2\pi}_{-2\pi}\frac{\sec^2{t}}{({\sqrt{a+1}})^2+\tan^2{t}}\,dt \ \ \ \ (\text{by substituting $t =2x$)}\\ &= \frac{1}{2}\int^{0}_{0}\frac{du}{({\sqrt{a+1}})^2+u^2} \ \ \ \ (\text{by substituting $u =\tan{t}$)}\\ &=0. \end{align} $$
But when I use some CAS and graph the function, I realise that for $a>-1$ area under the curve $y=1/({a\cos^2{2x}+1})$ remains positive. Even if I use indefinite integral of $y$, I get the same answer as you can see here:
Continuing from the last step: $$ \begin{align} \int{y}\,dx &= \frac{1}{2}\int\frac{du}{({\sqrt{a+1}})^2+u^2}\\ &=\frac{1}{2\sqrt{a+1}}\tan^{-1} \left( {\frac{u}{\sqrt{a+1}}} \right)\\ &=\frac{1}{2\sqrt{a+1}}\tan^{-1} \left( {\frac{\tan{t}}{\sqrt{a+1}}} \right)\\ \text{or} \int{y}\,dx \ &=\frac{1}{2\sqrt{a+1}}\tan^{-1} \left( {\frac{\tan{2x}}{\sqrt{a+1}}} \right) \end{align} $$
Therefore, $$ I = \frac{1}{2\sqrt{a+1}}\left[{\tan^{-1} \left( {\frac{\tan{2x}}{\sqrt{a+1}}} \right)}\right]_{-\pi}^{\pi} = 0. $$
I fail to grasp what's wrong with my way.