1192. Find the integrals:
1) $\int \frac{\cos x}{\sqrt{3+\sin ^{2} x}} d x$
2) $\int \sin ^{5} x d x$
3) $\int \frac{1}{3 \tan^{2} x+5} \cdot \frac{d x}{\cos ^{2} x}$
4) $\int \cot^{6} x d x$
Solution. 1) We use the substitution (1):

$$
\begin{gathered}
\int \frac{\cos x}{\sqrt{3+\sin ^{2} x}} d x=\int \frac{d t}{\sqrt{3+t^{2}}}=\ln \left(t+\sqrt{3+t^{2}}\right)+C= \\
=\ln \left(\sin x+\sqrt{3+\sin ^{2} x}\right)+C
\end{gathered}
$$

2) We apply the substitution (2):

$\int \sin ^{5} x d x=\int\left(1-\cos ^{2} x\right)^{2} \sin x d x=-\int\left(1-t^{2}\right)^{2} d t=-\int\left(1-2 t^{2}+t^{4}\right) d t=\frac{2 t^{3}}{3}-\frac{t^{5}}{5}-t+C=\frac{2}{3} \cos ^{3} x-\frac{1}{5} \cos ^{5} x-\cos x+C$.

3) We apply the substitution (3):

$$
\begin{aligned}
& \int \frac{1}{3 \operatorname{tg}^{2} x+5} \cdot \frac{d x}{\cos ^{2} x}=\int \frac{d t}{3 t^{2}+5}=\frac{1}{3} \int \frac{d t}{t^{2}+\frac{5}{3}}= \\
= & \frac{\sqrt{3}}{3 \sqrt{5}} \operatorname{arctg} \frac{t \sqrt{3}}{\sqrt{5}}+C=\frac{1}{\sqrt{15}} \operatorname{arctg} \frac{\sqrt{3} \operatorname{tg} x}{\sqrt{5}}+C .
\end{aligned}
$$

4) We use the substitution (4):

$$
\begin{gathered}
\int \operatorname{ctg}^{6} x d x=-\int t^{6} \frac{d t}{t^{2}+1}=-\int\left(t^{4}-t^{2}+1-\frac{1}{t^{2}+1}\right) d t= \\
=\frac{t^{3}}{3}-\frac{t^{5}}{5}-t-\operatorname{arctg} t+C=\frac{1}{3} \operatorname{ctg}^{3} x-\frac{1}{5} \operatorname{ctg}^{5} x- \\
-\operatorname{ctg} x-x+C \quad(0<x<\pi) .
\end{gathered}
$$