求不定积分:∫ [(sinx)^2]dx= ∫ [(sinx)^3]dx= ∫ [(sinx)^4]dx=
1个回答

∫ sin²x dx = ∫ (1 - cos2x)/2 dx = [x - (1/2)sin2x]/2 + C

∫ sin³x dx = ∫ (cos²x - 1) dcosx = (1/3)cos³x - cosx + C

∫ sin⁴x dx = ∫ [(1 - cos2x)/2]² dx = (1/4)∫ (1 - 2cos2x + cos²2x) dx

= (x - sin2x)/4 + (1/4)∫ (1 + cos4x)/2 dx

= (x - sin2x)/4 + (x + 1/4 • sin4x)/8 + C

= 3x/8 - (1/4)sin2x + (1/32)sin4x + C