As #d(cosx) = -sinx dx# we can write the integral as:
#int sin^4xdx = int sin^3 sinx dx = - int sin^3 d(cosx)#
and integrate by parts:
#int sin^4xdx = - sin^3x cosx + 3 int sin^2x cos^2x dx#
Now applying the identity:
#cos^2x = 1 - sin^2x#
#int sin^4xdx = - sin^3x cosx + 3 int sin^2x (1-sin^2x) dx#
and as the integral is linear:
#int sin^4xdx = - sin^3x cosx + 3 int sin^2xdx -3intsin^4x dx#
we have now the integral on both sides and we can solve for it:
#int sin^4xdx = - (sin^3x cosx )/4+ 3/4 int sin^2xdx #
We can now apply the same process for the integral:
# int sin^2xdx = -int sinx (dcosx) = -sinxcosx + int cos^2xdx = -sinxcosx + int (1-sin^2x)dx = -sinxcosx + int dx - int sin^2xdx#
and we get:
# int sin^2xdx = -(sinxcosx)/2 +x/2+C'#
Putting it together:
#int sin^4xdx = - (sin^3x cosx )/4-3/8sinxcosx +3/8x+C#
Note that you can write this result in an interesting form: first we use the identity:
#2sinx cosx = sin2x#
#- (sin^3x cosx )/4-3/8sinxcosx +3/8x = - (sin^2x sin2x)/8 -3/16sin2x +3/8x#
than we use:
#sin^2x = (1-cos2x)/2#
#- (sin^2x sin2x)/8 -3/16sin2x +3/8x = - ((1-cos2x) sin2x)/16 -3/16sin2x +3/8x = 1/32sin4x - 1/4sin2x +3/8x#
