Use the trigonometric identity:
cos^2x = 1-sin^2xcos2x=1−sin2x
int sin^4xcos^2x dx = int sin^4x(1-sin^2x)dx∫sin4xcos2xdx=∫sin4x(1−sin2x)dx
and the linearity of the integral:
int sin^4xcos^2x dx = int sin^4xdx-int sin^6xdx∫sin4xcos2xdx=∫sin4xdx−∫sin6xdx
Solve now:
int sin^6xdx = int sin^5sinx dx∫sin6xdx=∫sin5sinxdx
integrating by parts:
int sin^6xdx = int sin^5 d(-cosx) ∫sin6xdx=∫sin5d(−cosx)
int sin^6xdx = -sin^5x cosx + 5 int sin^4 cos^2xdx ∫sin6xdx=−sin5xcosx+5∫sin4cos2xdx
int sin^6xdx = -sin^5x cosx + 5 int sin^4 (1-sin^2x)dx ∫sin6xdx=−sin5xcosx+5∫sin4(1−sin2x)dx
int sin^6xdx = -sin^5x cosx + 5 int sin^4 dx -5 int sin^6xdx ∫sin6xdx=−sin5xcosx+5∫sin4dx−5∫sin6xdx
int sin^6xdx = -(sin^5x cosx)/6 + 5/6 int sin^4 dx ∫sin6xdx=−sin5xcosx6+56∫sin4dx
And then:
int sin^4xdx = int sin^3 d(-cosx) ∫sin4xdx=∫sin3d(−cosx)
int sin^4xdx = -sin^3x cosx + 3 int sin^2 cos^2xdx ∫sin4xdx=−sin3xcosx+3∫sin2cos2xdx
int sin^4xdx = -sin^3x cosx + 3 int sin^2 (1-sin^2x)dx ∫sin4xdx=−sin3xcosx+3∫sin2(1−sin2x)dx
int sin^4xdx = -sin^3x cosx + 3 int sin^2 dx -3 int sin^6xdx ∫sin4xdx=−sin3xcosx+3∫sin2dx−3∫sin6xdx
int sin^4xdx = -(sin^4x cosx)/4+ 3/4 int sin^2 dx ∫sin4xdx=−sin4xcosx4+34∫sin2dx
Finally:
int sin^2 dx = int (1-cos2x)/2 dx = x/2 -sin(2x)/4+C∫sin2dx=∫1−cos2x2dx=x2−sin(2x)4+C
int sin^2 dx = (x-sinxcosx)/2+C∫sin2dx=x−sinxcosx2+C
Putting partial results together:
int sin^4xcos^2x dx = int sin^4xdx-int sin^6xdx∫sin4xcos2xdx=∫sin4xdx−∫sin6xdx
int sin^4xcos^2x dx = int sin^4xdx +(sin^5x cosx)/6 - 5/6 int sin^4 dx ∫sin4xcos2xdx=∫sin4xdx+sin5xcosx6−56∫sin4dx
int sin^4xcos^2x dx = +(sin^5x cosx)/6 + 1/6 int sin^4 dx ∫sin4xcos2xdx=+sin5xcosx6+16∫sin4dx
int sin^4xcos^2x dx = +(sin^5x cosx)/6 -(sin^4x cosx)/24+3/48(x-sinxcosx)+C∫sin4xcos2xdx=+sin5xcosx6−sin4xcosx24+348(x−sinxcosx)+C
int sin^4xcos^2x dx = (cosx(8sin^5x-2sin^3x-3sinx))/48 +(3x)/48+C∫sin4xcos2xdx=cosx(8sin5x−2sin3x−3sinx)48+3x48+C
