As d(cosx) = -sinx dxd(cosx)=−sinxdx we can write the integral as:
int sin^4xdx = int sin^3 sinx dx = - int sin^3 d(cosx)∫sin4xdx=∫sin3sinxdx=−∫sin3d(cosx)
and integrate by parts:
int sin^4xdx = - sin^3x cosx + 3 int sin^2x cos^2x dx∫sin4xdx=−sin3xcosx+3∫sin2xcos2xdx
Now applying the identity:
cos^2x = 1 - sin^2xcos2x=1−sin2x
int sin^4xdx = - sin^3x cosx + 3 int sin^2x (1-sin^2x) dx∫sin4xdx=−sin3xcosx+3∫sin2x(1−sin2x)dx
and as the integral is linear:
int sin^4xdx = - sin^3x cosx + 3 int sin^2xdx -3intsin^4x dx∫sin4xdx=−sin3xcosx+3∫sin2xdx−3∫sin4xdx
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 ∫sin4xdx=−sin3xcosx4+34∫sin2xdx
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∫sin2xdx=−∫sinx(dcosx)=−sinxcosx+∫cos2xdx=−sinxcosx+∫(1−sin2x)dx=−sinxcosx+∫dx−∫sin2xdx
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
