Найти интеграл: \( \int \sin^5{x} dx .\)

Решение

\( \int \sin^5{x} dx = \int \sin^4{x} \sin{x} = \begin{vmatrix}
t = \cos{x}\\
t' = -\sin{x}\\
dt = -\sin{x} dt\\
dx = \frac{dx}{-\sin{x}}\\
\end{vmatrix} = \int \sin^4{x}\sin{x} \frac{-1}{\sin{x}}dt = \)

\(= \sin^4{x}dt = \int -(\sin^2{x})^2 dt = \int -(1-\cos^2{x})^2dt = \)

\(= - (1-t^2)^2 dt = \int -(1-2t^2+t^4)dt = \int (-q+2t^2-t^4)dt = -t+\frac{2t^3}{3}+\frac{t^3}{3}+C= \)

\(=-\cos{x} + \frac{2\cos^3{x}}{3} + \frac{\cos^5{x}}{5} +C . \)

Ответ: \(-\cos{x} + \frac{2\cos^3{x}}{3} + \frac{\cos^5{x}}{5} +C \).