How to solve ∫ (sin^4x)(cos^2x) dx ?

1 Answer
Sep 30, 2015

int sin^4x*cos^2x dx=1/16*(x-sin(4x)/4-(sin^3(2x))/3)+csin4xcos2xdx=116(xsin(4x)4sin3(2x)3)+c

Explanation:

intsin^4x*cos^2x dx=int(sin^2x)*(sin^2x*cos^2x)dxsin4xcos2xdx=(sin2x)(sin2xcos2x)dx

=int(1/2) * (1-cos2x) * ((sin2x)/2)^2dx=(12)(1cos2x)(sin2x2)2dx

We have

cos2x=1-2sin^2xcos2x=12sin2x

sin^2x=(1-cos2x)/2sin2x=1cos2x2

and

sin2x=2sinx*cosxsin2x=2sinxcosx

The integral becomes

=int(1/2) * (1-cos2x) * ((sin2x)/2)^2dx=(12)(1cos2x)(sin2x2)2dx

=1/8(int(1-cos2x)*sin^2(2x))dx=18((1cos2x)sin2(2x))dx

=1/8(int(sin^2(2x)-cos2x*sin^2(2x))dx=18((sin2(2x)cos2xsin2(2x))dx

=1/8(int(1/2)*2sin^2(2x)dx-int1/2*2cos2x*sin^2(2x)dx)=18((12)2sin2(2x)dx122cos2xsin2(2x)dx)

=1/8(int(1/2)*(1-cos(4x))dx-int1/2*sin^2(2x)d(sin2x))=18((12)(1cos(4x))dx12sin2(2x)d(sin2x))

=1/16(int(1-cos(4x))dx-intsin^2(2x)d(sin2x))=116((1cos(4x))dxsin2(2x)d(sin2x))

d(sin2x)/dx=2cos2xdsin2xdx=2cos2x

d(sin2x)=2cos2x*dxd(sin2x)=2cos2xdx

=1/16(x-sin(4x)/4-sin^3(2x)/3)+c=116(xsin(4x)4sin3(2x)3)+c