Show that integration of cos^4 x sin² x dx = 1/16 [ x - (sin4x)/4 + (sin^3 2x)/3 ] +c ?

1 Answer
Jothi S.
Dec 10, 2015

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

Explanation:

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

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