See question Evaluating $\int P(\sin x, \cos x) \text{d}x$.
The universal standard substitution to evaluate an integral of a rational fraction in $\sin \theta,\cos \theta$, i.e. a rational fraction of the form
$$R(\sin \theta,\cos \theta)=\frac{P(\sin \theta,\cos \theta)}{Q(\sin \theta,\cos \theta)},$$
where $P,Q$ are polynomials in $\sin \theta,\cos \theta$ is a trigonometric substitution known as the Weierstrass substitution
$$
\begin{equation*}
\tan \frac{\theta }{2}=t,\qquad\theta =2\arctan t,\qquad d\theta =\frac{2}{1+t^{2}}dt
\end{equation*},
$$
which converts the integrand into a rational function in $t$. We know from trigonometry (see this answer) that
$$\cos \theta =\frac{1-\tan ^{2}\frac{\theta }{2}}{1+\tan ^{2}\frac{
\theta}{2}}=\frac{1-t^2}{1+t^2},\qquad \sin \theta =\frac{2\tan \frac{\theta }{2}}{1+\tan ^{2}
\frac{\theta }{2}}=\frac{2t}{1+t^2}.$$
Applying this substitution to e.g. the first integral, we get
$$
\begin{eqnarray*}
I &=&\int \left( \frac{\cos \theta }{1+\sin ^{2}\theta }\right) ^{2}d\theta
=\int \left( \frac{\cos \theta }{2-\cos ^{2}\theta }\right) ^{2}d\theta \\
&=&\int \frac{2\left( 1+t^{2}\right) \left( t^{2}-1\right) ^{2}}{\left(
1+6t^{2}+t^{4}\right) ^{2}}dt.
\end{eqnarray*}
$$
[Edited and added.] The integrand can be expanded as
$$
\begin{eqnarray*}
\frac{2\left( 1+t^{2}\right) \left( t^{2}-1\right) ^{2}}{\left(
1+6t^{2}+t^{4}\right) ^{2}} &=&\frac{16+80t^{2}}{\left(
1+6t^{2}+t^{4}\right) ^{2}}+\frac{2t^{2}-14}{1+6t^{2}+t^{4}} \\
&=&\frac{1}{4}\frac{4-3\sqrt{2}}{t^{2}+3-2\sqrt{2}}+\frac{5\sqrt{2}-7}{
\left( t^{2}+3-2\sqrt{2}\right) ^{2}}\\&+&\frac{1}{4}\frac{4+3\sqrt{2}}{t^{2}+3+2
\sqrt{2}}-\frac{7+5\sqrt{2}}{\left( t^{2}+3+2\sqrt{2}\right) ^{2}}.
\end{eqnarray*}
$$
With the help of SWP I evaluated
$$
\begin{eqnarray*}
\int \frac{1}{4}\frac{4\mp 3\sqrt{2}}{t^{2}+3-2\sqrt{2}}dt &=&\frac{1}{4}
\frac{4\pm 3\sqrt{2}}{\sqrt{2}-1}\arctan \frac{t}{\sqrt{2}-1}, \\
\int \frac{-7+5\sqrt{2}}{\left( t^{2}+3-2\sqrt{2}\right) ^{2}}dt &=&\frac{
(7-5\sqrt{2})t}{2\left( -3+2\sqrt{2}\right) \left( t^{2}+3-2\sqrt{2}\right) }\\
&+&\frac{7-5\sqrt{2}}{2\left( -3+2\sqrt{2}\right) \left( \sqrt{2}-1\right) }
\arctan \frac{t}{\sqrt{2}-1}, \\
-\int \frac{7+5\sqrt{2}}{\left( t^{2}+3+2\sqrt{2}\right) ^{2}}dt &=&-\frac{
(7+5\sqrt{2})t}{2\left( 3+2\sqrt{2}\right) \left( t^{2}+3+2\sqrt{2}\right) }\\&-&
\frac{7+5\sqrt{2}}{2\left( 3+2\sqrt{2}\right) \left( \sqrt{2}+1\right) }
\arctan \frac{t}{\sqrt{2}+1}.
\end{eqnarray*}
$$
As a consequence the integral $I$ can be written as
$$
\begin{eqnarray*}
I &=&\frac{1}{4}\frac{-10+7\sqrt{2}}{-5\sqrt{2}+7}\arctan \frac{t}{\sqrt{2}-1}+\frac{1}{4}\frac{10+7\sqrt{2}}{7+5\sqrt{2}}\arctan \frac{t}{\sqrt{2}+1}\\&-&\frac{t^{3}-t}{t^{4}+6t^{2}+1}+C, \\
&=&\frac{1}{4}\frac{-10+7\sqrt{2}}{-5\sqrt{2}+7}\arctan \frac{\tan \frac{
\theta }{2}}{\sqrt{2}-1}+\frac{1}{4}\frac{10+7\sqrt{2}}{7+5\sqrt{2}}\arctan
\frac{\tan \frac{\theta }{2}}{\sqrt{2}+1}\\&-&\frac{ \tan ^{3}\frac{\theta }{2}
-\tan \frac{\theta }{2}}{ \tan^{4} \frac{\theta }{2}
+6 \tan ^{2}\frac{\theta }{2} +1}+C.
\end{eqnarray*}
$$