Antworten:
# = 1/16 (x-sin (4x) / 4 + sin ^ 3 (2x) / 3) #
Erläuterung:
#int (cos ^ 4 (x) sin ^ 2 (x)) dx = int ((1 + cos (2x)) / 2) ^ 2 ((1-cos (2x)) / 2) dx #
Formel verwenden
# cos ^ 2 (x) = (1 + cos (2x)) / 2 #
# sin ^ 2 (2x) = (1-cos (2x)) / 2 #
#int ((1 + cos (2x)) / 2) ^ 2 ((1-cos (2x)) / 2) dx #
# = int ((1 + cos ^ 2 (2x) + 2cos (2x)) (1-cos (2x))) / 8dx #
# = int ((1 + cos ^ 2 (2x) + 2cos (2x) -cos (2x) -cos ^ 3 (2x) -2cos ^ 2 (2x)) / 8) dx #
#int (1 + cos (2x) -cos ^ 2 (2x) -cos ^ 3 (2x)) / 8dx #
# 1/8 (int (dx) + int cos (2x) dx-int (cos ^ 2 (2x) dx-int (cos ^ 3 (dx) #
#int cos ^ 2 (2x) dx = int (1 + cos (4x)) / 2dx #=# x / 2 + sin (4x) / 8 #
# intcos ^ 3 (2x) dx = int (1-sin ^ 2 (2x)) cos (2x) dx #
# = int cos (2x) -sin ^ 2 (2x) cos (2x) dx = sin (2x) / 2-sin ^ 3 (2x) / 6 #
# 1/8 (int (dx) + int cos (2x) dx-int (cos ^ 2 (2x) dx-int (cos ^ 3 (dx) #
=# 1/8 (x + sin (2x) / 2-x / 2-sin (4x) / 8-sin (2x) / 2 + sin ^ 3 (2x) / 6) #
# = 1/16 (x-sin (4x) / 4 + sin ^ 3 (2x) / 3) #
![Zeigen Sie, dass die Integration von cos ^ 4 x sin² x dx = 1/16 [x - (sin4x) / 4 + (sin ^ 3 2x) / 3] + c?](https://img.go-homework.com/img/calculus/show-that-integration-of-cos4-x-sin-x-dx-1/16-x-sin4x/4-sin3-2x/3-c-.png "Zeigen Sie, dass die Integration von cos ^ 4 x sin² x dx = 1/16 [x - (sin4x) / 4 + (sin ^ 3 2x) / 3] + c?")
