 $$
            \int\frac1{\sin^2 x \cos^2 x}\dx
      $$

      \textbf{�e�en�}
      Pou�ijeme rovnost $1 = \sin^2 x + \cos^2 x$:
      \begin{multline*}
            \int\frac1{\sin^2 x \cos^2 x}\dx = \int \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x}\dx
            =\int\left(\frac{\cos^2 x}{\sin^2 x \cos^2 x} + \frac{\sin^2 x}{\sin^2 x \cos^2 x}\right)\dx =\\
            =\int\left(\frac 1{\cos^2 x} + \frac 1{\sin^2 x}\right)\dx \stackrel c= \tg x - \cotg x.
      \end{multline*}
