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?
 A: The integrand is a even funtion and has period $2\pi
 $. We recall that if we have a function of period $T
 $ then for all $a\in\mathbb{R}
 $ holds $$\int_{0}^{T}f\left(x\right)dx=\int_{a}^{T+a}f\left(x\right)dx
 $$ then $$\begin{align}
 -\int_{0}^{2\pi}\frac{\cos\left(t\right)\left(2\sin^{2}\left(t\right)+\cos^{2}\left(t\right)\right)}{\cos^{4}\left(t\right)+\sin^{2}\left(t\right)}dt= & -\int_{.-\pi}^{\pi}\frac{\cos\left(t\right)\left(2\sin^{2}\left(t\right)+\cos^{2}\left(t\right)\right)}{\cos^{4}\left(t\right)+\sin^{2}\left(t\right)}dt \\
 \stackrel{t=u-\pi}{=} & -\int_{0}^{2\pi}\frac{\cos\left(u-\pi\right)\left(2\sin^{2}\left(u-\pi\right)+\cos^{2}\left(u-\pi\right)\right)}{\cos^{4}\left(u-\pi\right)+\sin^{2}\left(u-\pi\right)}dt \\
 = & \int_{0}^{2\pi}\frac{\cos\left(t\right)\left(2\sin^{2}\left(t\right)+\cos^{2}\left(t\right)\right)}{\cos^{4}\left(t\right)+\sin^{2}\left(t\right)}dt
\end{align}
 $$ hence $$\int_{0}^{2\pi}\frac{\cos\left(t\right)\left(2\sin^{2}\left(t\right)+\cos^{2}\left(t\right)\right)}{\cos^{4}\left(t\right)+\sin^{2}\left(t\right)}dt=0.
 $$
A: \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 
$$I=-2\int_{0}^{1}{\frac{\ 1+{{u}^{2}}}{{{(1-{{u}^{2}})}^{2}}+{{u}^{2}}}du}+2\int_{0}^{1}{\frac{\ 1+{{u}^{2}}}{{{(1-{{u}^{2}})}^{2}}+{{u}^{2}}}du}=0$$
A: HINT: Consider $\int_{-\pi/2}^{\pi/2} f(t)\,dt + \int_{\pi/2}^{3\pi/2} f(t)\,dt$. In particular, what is $f(t+\pi)$ compared to $f(t)$?
