Pokazati da integracija cos ^ 4 x sin² x dx = 1/16 [x - (sin4x) / 4 + (sin ^ 3 2x) / 3] + c?

Odgovor:

# = 1/16 (x-sin (4x) / 4 + sin ^ 3 (2 x) / 3) *

Obrazloženje:

#int (cos ^ 4 (x) ^ 2 sin (x)) dx = int ((1 + cos (2x)) / 2) ^ 2 ((1-cos (2 x)) / 2) dx #

Koristeći formulu

# cos ^ 2 (x) = (1 + cos (2x)) / 2 #

# Sin ^ 2 (2 x) = (1-cos (2 x)) / 2 #

#int ((1 + cos (2x)) / 2) ^ 2 ((1-cos (2 x)) / 2) dx #

# = Int ((1 + cos ^ 2 (2x) + 2cos (2x)) (1-cos (2 x))) / 8dx #

# = Int ((1 + cos ^ 2 (2x) + 2cos (2x) -cos (2x) -cos ^ 3 (2 x) -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 (2 x) dx = int (1-sin ^ 2 (2x)) cos (2 x) 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 (2 x) / 2x / 2-sin (4x) / 8-sin (2 x) / 2 + sin ^ 3 (2 x) / 6) #

# = 1/16 (x-sin (4x) / 4 + sin ^ 3 (2 x) / 3) *