Use the trigonometric identity:
#cos^2x = 1-sin^2x#
#int sin^4xcos^2x dx = int sin^4x(1-sin^2x)dx#
and the linearity of the integral:
#int sin^4xcos^2x dx = int sin^4xdx-int sin^6xdx#
Solve now:
#int sin^6xdx = int sin^5sinx dx#
integrating by parts:
#int sin^6xdx = int sin^5 d(-cosx) #
#int sin^6xdx = -sin^5x cosx + 5 int sin^4 cos^2xdx #
#int sin^6xdx = -sin^5x cosx + 5 int sin^4 (1-sin^2x)dx #
#int sin^6xdx = -sin^5x cosx + 5 int sin^4 dx -5 int sin^6xdx #
#int sin^6xdx = -(sin^5x cosx)/6 + 5/6 int sin^4 dx #
And then:
#int sin^4xdx = int sin^3 d(-cosx) #
#int sin^4xdx = -sin^3x cosx + 3 int sin^2 cos^2xdx #
#int sin^4xdx = -sin^3x cosx + 3 int sin^2 (1-sin^2x)dx #
#int sin^4xdx = -sin^3x cosx + 3 int sin^2 dx -3 int sin^6xdx #
#int sin^4xdx = -(sin^4x cosx)/4+ 3/4 int sin^2 dx #
Finally:
# int sin^2 dx = int (1-cos2x)/2 dx = x/2 -sin(2x)/4+C#
# int sin^2 dx = (x-sinxcosx)/2+C#
Putting partial results together:
#int sin^4xcos^2x dx = int sin^4xdx-int sin^6xdx#
#int sin^4xcos^2x dx = int sin^4xdx +(sin^5x cosx)/6 - 5/6 int sin^4 dx #
#int sin^4xcos^2x dx = +(sin^5x cosx)/6 + 1/6 int sin^4 dx #
#int sin^4xcos^2x dx = +(sin^5x cosx)/6 -(sin^4x cosx)/24+3/48(x-sinxcosx)+C#
#int sin^4xcos^2x dx = (cosx(8sin^5x-2sin^3x-3sinx))/48 +(3x)/48+C#
