How to calculate this integral in one variable I want calculate this integral
$$ \int_0^{2\pi} - \frac{\cos t \; ( 2 (\sin t)^2 + (\cos t)^2)}{(\cos t)^4 + (\sin t)^2} \, dt $$
Can I use an opportunity substitution?

\begin{align}
  & I=\int_{0}^{2\pi }{-}\frac{\cos t\ (2{{\sin }^{2}}t+{{\cos }^{2}}t)}{{{\cos }^{4}}t+{{\sin }^{2}}t}dt=-2\int_{0}^{\pi }{\frac{\cos t\ (1+{{\sin }^{2}}t)}{{{(1-{{\sin }^{2}}t)}^{2}}+{{\sin }^{2}}t}dt} \\ 
 & I=-2\int_{0}^{\frac{\pi }{2}}{\frac{\cos t\ (1+{{\sin }^{2}}t)}{{{(1-{{\sin }^{2}}t)}^{2}}+{{(\sin t)}^{2}}}dt}-2\int_{\frac{\pi }{2}}^{\pi }{\frac{\cos t\ (1+{{\sin }^{2}}t)}{{{(1-{{\sin }^{2}}t)}^{2}}+{{(\sin t)}^{2}}}dt} \\ 
\end{align}
let $u=\sin t$, we have 
$$I=-2\int_{0}^{1}{\frac{\ 1+{{u}^{2}}}{{{(1-{{u}^{2}})}^{2}}+{{u}^{2}}}du}+2\int_{0}^{1}{\frac{\ 1+{{u}^{2}}}{{{(1-{{u}^{2}})}^{2}}+{{u}^{2}}}du}=0$$
