1193. Find the integrals:
1) $\int \sin ^{2} x \cos ^{4} x d x ;$ 2) $\int \cos ^{4} x d x ;$ 3) $\int \frac{d x}{2+3 \cos x}$.
Solution. 1) Applying the formulas from point 3, we get

$$
\begin{gathered}
I=\int \sin ^{2} x \cos ^{4} x d x=\frac{1}{4} \int \sin ^{2} 2 x \frac{1+\cos 2 x}{2} d x= \\
=\frac{1}{8} \int \sin ^{2} 2 x d x+\frac{1}{8} \int \sin ^{2} 2 x \cos 2 x d x=\frac{1}{16} \int(1-\cos 4 x) d x+ \\
+\frac{1}{16} \int t^{2} d t
\end{gathered}
$$

where $t=\sin 2 x ; \quad d t=2 \cos 2 x d x$.

Finally, we find

$$
\begin{gathered}
I=\frac{x}{16}-\frac{\sin 4 x}{64}+\frac{\sin ^{3} 2 x}{48}+C . \\
2) \int \cos ^{4} x d x=\int\left(\frac{1+\cos 2 x}{2}\right)^{2} d x=\frac{1}{4} \int(1+2 \cos 2 x+ \\
+\cos ^{2} 2 x) d x=\frac{1}{4} \int d x+\frac{1}{2} \int \cos 2 x d x+\frac{1}{8} \int(1+\cos 4 x) d x= \\
=\frac{x}{4}+\frac{\sin 2 x}{4}+\frac{x}{8}+\frac{\sin 4 x}{32}+C=\frac{3 x}{8}+\frac{\sin 2 x}{4}+\frac{\sin 4 x}{32}+C .
\end{gathered}
$$

3) To compute this integral, we use the universal substitution $t=\operatorname{tg} \frac{x}{2}$, from which $d t=\frac{1}{2} \sec ^{2} \frac{x}{2} d x$;

$$
\begin{aligned}
& \int \frac{d x}{2+3 \cos x}=\int \frac{d x}{2+3 \cos ^{2} \frac{x}{2}-3 \sin ^{2} \frac{x}{2}}=\int \frac{\sec ^{2} \frac{x}{2} \cdot d x}{2 \sec ^{2} \frac{x}{2}+3-3 \operatorname{tg}^{2} \frac{x}{2}}= \\
& =\int \frac{\sec ^{2} \frac{x}{2} d x}{2\left(\operatorname{tg}^{2} \frac{x}{2}+1\right)+3-3 \operatorname{tg}^{2} \frac{x}{2}}=\int \frac{\sec ^{2} \frac{x}{2} d x}{5-\operatorname{tg}^{2} \frac{x}{2}}=-\int \frac{2 d t}{t^{2}-5}= \\
& =-\frac{2}{2 \sqrt{5}} \ln \left|\frac{t-\sqrt{5}}{t+\sqrt{5}}\right|+C=\frac{\sqrt{5}}{5} \ln \left|\frac{\operatorname{tg} \frac{x}{2}+\sqrt{5}}{\operatorname{tg} \frac{x}{2}-\sqrt{5}}\right|+C
\end{aligned}
$$

Find the integrals: