Example 1.9. Find $\int \frac{\sin 2 x d x}{\sin ^{4} x+\cos ^{4} x}$.
Solution. Since the integrand is an even function of $\sin x$ and $\cos x$:

$$
\begin{aligned}
R(\sin x, \cos x) & =\frac{2}{\sin ^{4} x+\cos ^{4} x} \\
R(-\sin x,-\cos x) & =\frac{2(-\sin x)(-\cos x)}{(-\sin x)^{4}+(-\cos x)^{4}}= \\
& =\frac{2 \sin x \cos x}{\sin ^{4} x+\cos ^{4} x}=R(\sin x, \cos x)
\end{aligned}
$$

we can apply the substitution $t=\tan x$. In this case,

$$
\begin{gathered}
\sin ^{2} x=\frac{\tan ^{2} x}{1+\tan ^{2} x}=\frac{t^{2}}{1+t^{2}}, \quad \cos ^{2} x=\frac{1}{1+\tan ^{2} x}=\frac{1}{1+t^{2}} \\
x=\operatorname{arcctg} t, \quad d x=\frac{d t}{1+t^{2}}
\end{gathered}
$$

Then

$$
\begin{aligned}
& \int \frac{\sin 2 x d x}{\sin ^{4} x+\cos ^{4} x}=2 \int \frac{\sin x \cos x d x}{\sin ^{4} x+\cos ^{4} x}=2 \int \frac{\tan x \cos ^{2} x d x}{\sin ^{4} x+\cos ^{4} x}= \\
= & 2 \int \frac{t /(1+t)^{2} d t /\left(1+t^{2}\right)}{t^{4} /\left(1+t^{2}\right)^{2}+1 /\left(1+t^{2}\right)^{2}}=2 \int \frac{t d t}{1+t^{4}}=\int \frac{d t^{2}}{1+\left(t^{2}\right)^{2}}= \\
= & \int \frac{d z}{1+z^{2}}=\operatorname{arctg} z+C=\operatorname{arctg} t^{2}+C=\operatorname{arctg}\left(\tan ^{2} x\right)+C
\end{aligned}
$$