How to solve ∫ (sin^4x)(cos^2x) dx ?
1 Answer
Explanation:
#intsin^4x*cos^2x dx=int(sin^2x)*(sin^2x*cos^2x)dx#
#=int(1/2) * (1-cos2x) * ((sin2x)/2)^2dx#
We have
#cos2x=1-2sin^2x#
#sin^2x=(1-cos2x)/2#
and
#sin2x=2sinx*cosx#
The integral becomes
#=int(1/2) * (1-cos2x) * ((sin2x)/2)^2dx#
#=1/8(int(1-cos2x)*sin^2(2x))dx#
#=1/8(int(sin^2(2x)-cos2x*sin^2(2x))dx#
#=1/8(int(1/2)*2sin^2(2x)dx-int1/2*2cos2x*sin^2(2x)dx)#
#=1/8(int(1/2)*(1-cos(4x))dx-int1/2*sin^2(2x)d(sin2x))#
#=1/16(int(1-cos(4x))dx-intsin^2(2x)d(sin2x))#
#d(sin2x)/dx=2cos2x#
#d(sin2x)=2cos2x*dx#