Найти интегралы:
\( \int \sin^2{x}\cos{x} dx \)

Решение

\( \int \sin^2{x}\cos{x} dx = \int (1-\cos^2{x})\cos{x} dx = \int (\cos{x} - \cos^3{x})dx = \sin{x} - \int \cos^3{x} dx + C = \)

\( \begin{vmatrix}\\
\int \cos^3{x} dx = \int \cos^2{x}\cos{x} dx = \int (1-\sin^2{x})\cos{x} dx = \begin{vmatrix}
t = \sin{x}\\
t' = \cos{x}\\
dt = dx\cos{x}\\
dx = \frac{dt}{\cos{x}} \end{vmatrix} =\\
= \int (1-t^2) \frac{\cos {x}}{\cos{x}} dt = \int (1-t^2) dt = t - \frac{t^3}{3} + c = \sin{x} - \frac{\sin^3{x}}{3} + C \end{vmatrix}\)

\( = \sin{x} - \sin{x} + \frac{\sin^3{x}}{3} + C = \frac{\sin^3{x}}{3} + C \)

Ответ: \( \frac{\sin^3{x}}{3} + C \).