Example 1.8. Find $\int\left(\frac{1-\cos x}{1+\cos x}\right)^{2} \frac{d x}{3 \sin x+4 \cos x+5}$.
Solution. Applying the universal trigonometric substitution, we get

$$
\begin{aligned}
& \int\left(\frac{1-\cos x}{1+\cos x}\right)^{2} \frac{d x}{3 \sin x+4 \cos x+5}= \\
= & \int \frac{\left(1-\frac{1-t^{2}}{1+t^{2}}\right)^{2}}{\left(1+\frac{1-t^{2}}{1+t^{2}}\right)^{2}} \frac{2 d t}{\left(1+t^{2}\right)\left(\frac{6 t}{1+t^{2}}+\frac{4\left(1-t^{2}\right)}{1+t^{2}}+5\right)}= \\
= & 2 \int \frac{t^{4} d t}{6 t+4-4 t^{2}+5+5 t^{2}}=2 \int \frac{t^{4} d t}{t^{2}+6 t+9}=2 \int \frac{t^{4} d t}{(t+3)^{2}} .
\end{aligned}
$$

The rational fraction whose integral we need to find is improper. Therefore, before decomposing it into the sum of the simplest fractions, we need to extract the integer part. To do this, we divide the numerator by the denominator:

$$
\begin{aligned}
& t^{4} \\
& t^{4}-6 t^{3}+9 t^{2} \\
& \left\lvert\, \begin{array}{l}
t^{2}+6 t+9 \\
t^{2}-6 t+27
\end{array}\right. \\
& -6 t^{3}-9 t^{2} \\
& \frac{-6 t^{2}-36 t^{2}-54 t}{27 t^{2}+54 t} \\
& \frac{27 t^{2}+162 t+243}{-108 t-243}
\end{aligned}
$$

Thus,

$$
\int \frac{t^{4}}{t^{2}+6 t+9}=t^{2}-6 t+27-\frac{108 t+243}{t^{2}+6 t+9}
$$

Considering this result, we have:

$$
\begin{gathered}
\int \frac{t^{4} d t}{(t+3)^{2}}=\int\left(t^{2}+6 t+9\right) \cdot d t-\int \frac{108 t+243}{(t+3)^{2}} d t= \\
=\frac{t^{3}}{3}-3 t^{2}+27 t-108 \int \frac{(t+3)}{(t+3)^{2}} d t+81 \int \frac{d t}{(t+3)^{2}}= \\
=\frac{t^{3}}{3}-3 t^{2}+27 t-108 \ln |t+3|-\frac{81}{t+3}+C
\end{gathered}
$$

Returning to the old variable, we get:

$$
\begin{aligned}
& \int\left(\frac{1-\cos x}{1+\cos x}\right)^{2} \frac{d x}{3 \sin x+4 \cos x+5}= \\
= & 2\left(\frac{1}{3} \operatorname{tg}^{3} \frac{x}{2}-3 \operatorname{tg}^{2} \frac{x}{2}+27 \operatorname{tg} \frac{x}{2}-108 \ln \left|\operatorname{tg} \frac{x}{2}+3\right|-\frac{81}{\operatorname{tg} \frac{x}{2}+3}\right)+C
\end{aligned}
$$

The universal trigonometric substitution often leads to cumbersome integrands and, as a consequence, to a large volume of computational work. Therefore, in some cases, other substitutions are used, which significantly simplify the finding of the integral compared to the situation when the universal trigonometric substitution is applied (in the same case).

Let us indicate some particular cases when the integral $\int R(\sin x \cos x) d x$ is rationalized using other substitutions, leading to less cumbersome calculations:

a) if $R(\sin x, \cos x)$ is an odd function with respect to $\sin x$, that is, $R(-\sin x, \cos x)=-R(\sin x, \cos x)$, then the integral (1.19) is rationalized by the substitution $t=\cos x$;

b) if $R(\sin x, \cos x)$ is an odd function with respect to $\cos x$, that is, $R(\sin x,-\cos x)=-R(\sin x, \cos x)$, then the integral (1.19) is rationalized by the substitution $t=\sin x$;

c) if $R(\sin x, \cos x)$ is an even function with respect to $\sin x$ and $\cos x$, that is, $R(-\sin x,-\cos x)=R(\sin x, \cos x)$, then the integral (1.19) is rationalized by the substitution $t=\operatorname{tg} x$.