Задача №1618

\[ \int\tg^5{x}dx =\left[\begin{aligned} & u=\tg{x};\\ & dx=\frac{du}{1+u^2}. \end{aligned}\right] =\int\frac{u^5du}{1+u^2} =\int\frac{\left(u^4-1+1\right)\cdot{u}}{1+u^2}du =\int\frac{\left(\left(u^2-1\right)\left(u^2+1\right)+1\right)\cdot{u}}{1+u^2}du=\\ =\int\left(\left(u^2-1\right)u+\frac{u}{u^2+1} \right) =\frac{u^4}{4}-\frac{u^2}{2}+\frac{1}{2}\ln\left(u^2+1\right)+C =\frac{\tg^4{x}}{4}-\frac{\tg^2{x}}{2}+\frac{1}{2}\ln\left(\tg^2{x}+1\right)+C=\\ =\frac{\tg^4{x}}{4}-\frac{\tg^2{x}}{2}+\frac{1}{2}\ln\frac{1}{\cos^2{x}}+C =\frac{\tg^4{x}}{4}-\frac{\tg^2{x}}{2}-\ln|\cos{x}|+C. \]