求∫(sinx)^4 dx ~!!!

Ans : ( 3x / 8 ) - ( sin2x / 4 ) + ( sin4x / 32 ) + C

Solutions :

cos2x = 1 - 2sin² x
sin² x = 1/2 ( 1 - cos2x )
sin^4 = 1/4 ( 1 - 2cos2x + cos² 2x )

∫(sinx)^4 dx

= 1/4 ∫( 1 - 2cos2x + cos² 2x )dx

= 1/4∫1dx - 1/2∫cos2x dx + 1/4∫cos² 2x dx
= x/4 - 1/4∫cos2x d2x + 1/8∫( cos4x + 1 )dx
= x/4 - ( sin2x )/4 + 1/32∫cos4x d4x + 1/8∫1dx
= x/4 - ( sin2x )/4 + ( sin4x / 32 ) + x/8
= ( 3x / 8 ) - ( sin2x / 4 ) + ( sin4x / 32 ) + C, where C is a constant.