182. $\int \tan x d x$
Solution. First, we transform the integrand: $\operatorname{tg} x=\frac{\sin x}{\cos x}$. Next, we find

$$
\int \lg x d x=\int \frac{\sin x d x}{\cos x}=\left|\begin{array}{c}
\cos x=z \\
-\sin x d x=d z \\
\sin x d x=-d z
\end{array}\right|=
$$

$$
=-\int \frac{d z}{z}=-\ln |z|+C=-\ln |\cos x|+C
$$