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)+c

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

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