Задача №2360

\[ \int\frac{dx}{\tg{x}\cdot\cos{2x}} =\int\frac{dx}{\tg{x}\cdot\left(\cos^2{x} - \sin^2{x}\right)} =\int\frac{\frac{dx}{\cos^2{x}}}{\tg{x}\cdot\left(1 - \tg^2{x}\right)} = \\ =\int\frac{d(\tg{x})}{\tg{x}\cdot\left(1 - \tg^2{x}\right)} =\left[ t=\tg{x} \right] =\int\frac{dt}{t\left(1-t^2\right)} =\int\left(\frac{1}{t} -\frac{1}{2}\cdot\frac{1}{t-1} - \frac{1}{2}\cdot\frac{1}{t+1}\right)dt =\\ =\ln\frac{|t|}{\sqrt{\left|1-t^2\right|}}+\ln|C| =\ln\frac{|Ct|}{\sqrt{\left|1-t^2\right|}}+\ln|C| =\ln\frac{|C\sin{x}|}{|\cos{x}|\sqrt{\left|1-\tg^2{x}\right|}} =\ln\frac{|C\sin{x}|}{\sqrt{|\cos{2x}|}} \]