Given ∫sin⁶ x dx

= ∫(sin² x)³ dx

= ∫{(1 - cos2x)/2}³ dx      [since cos2x = 1 - 2sin² x => sin² x = (1 - cos2x)/2]

= (1/8)*∫{(1 - cos2x)}³ dx

= (1/8)*∫(1 - cos³ 2x - 3cos2x + 3cos² 2x)} dx

= (1/8)*[∫dx - ∫cos³ 2x dx - 3∫cos2x dx + 3∫cos² 2x dx)}

= (1/8)*[ x - ∫cos³ 2x dx - (3/2)*sin2x + 3∫cos² 2x)} dx

Now ∫cos³ 2x dx = ∫cos2x * cos² 2x dx

                          = ∫cos2x * (1- sin² 2x) dx

Let sin2x = u

=> 2*cos2x dx = du

=> cos2x = du/2

∫cos2x * (1- sin² 2x) dx = (1/2)*∫(1- u² ) du

                                     = (1/2)*(u- u³ /3) 

                                     = (1/2)*{sin2x- (sin³ 2x)/3}

Again 3∫cos² 2x dx = (3/2)*∫(1 + cos4x) dx                   (since cos2x = 2cos² x - 1)

                              = (3/2)*(x + sin4x/4)

Now ∫ sin⁶ x dx = x/8 - (3/16)*sin2x - (1/16)*{sin2x - (sin³ 2x)/3} + (3/16)*{3 + (sin4x)/4} + c