Задача №1618
Условие
Найти интеграл \(\int\tg^5{x}dx\).
Решение
\[
\int\tg^5{x}dx
=\left[\begin{aligned}
& u=\tg{x};\\
& dx=\frac{du}{1+u^2}.
\end{aligned}\right]
=\int\frac{u^5du}{1+u^2}
=\int\frac{\left(u^4-1+1\right)\cdot{u}}{1+u^2}du
=\int\frac{\left(\left(u^2-1\right)\left(u^2+1\right)+1\right)\cdot{u}}{1+u^2}du=\\
=\int\left(\left(u^2-1\right)u+\frac{u}{u^2+1} \right)
=\frac{u^4}{4}-\frac{u^2}{2}+\frac{1}{2}\ln\left(u^2+1\right)+C
=\frac{\tg^4{x}}{4}-\frac{\tg^2{x}}{2}+\frac{1}{2}\ln\left(\tg^2{x}+1\right)+C=\\
=\frac{\tg^4{x}}{4}-\frac{\tg^2{x}}{2}+\frac{1}{2}\ln\frac{1}{\cos^2{x}}+C
=\frac{\tg^4{x}}{4}-\frac{\tg^2{x}}{2}-\ln|\cos{x}|+C.
\]
Ответ:
\(\frac{\tg^4{x}}{4}-\frac{\tg^2{x}}{2}-\ln|\cos{x}|+C\)