求 (1)∫sin2xcos3x dx (2)∫cosx(cosx/2) dx(...
(1):sin2x * cos3x = (1/2) * [sin(2x 3x) sin(2x-3x)]= (1/2) * [sin5x sin(-x)]= (1/2) * (sin5x - sinx)∫ sin2x * cos3x dx= (1/2)∫ (sin5x - sinx) dx= (1/2) * [(-1/5)cos5x - (-cosx)] c= (1/10) * (5cosx - cos5x) c(2):cosx * cos(x/2) = (1/2) * [cos(x x/2) cos(x-x/2)]= (1/2) * [cos(3x/2) cos(x...全部
(1):sin2x * cos3x = (1/2) * [sin(2x 3x) sin(2x-3x)]= (1/2) * [sin5x sin(-x)]= (1/2) * (sin5x - sinx)∫ sin2x * cos3x dx= (1/2)∫ (sin5x - sinx) dx= (1/2) * [(-1/5)cos5x - (-cosx)] c= (1/10) * (5cosx - cos5x) c(2):cosx * cos(x/2) = (1/2) * [cos(x x/2) cos(x-x/2)]= (1/2) * [cos(3x/2) cos(x/2)]∫ cosx * cos(x/2) dx= (1/2)∫ [cos(3x/2) cos(x/2)] dx= (1/2) * [2/3*sin(3x/2) 2sin(x/2)] c= sin(x/2) (1/3)sin(3x/2) c(3):sin5x * sin7x = (1/2) * [cos(5x-7x) - cos(5x 7x)]= (1/2) * [cos(-2x) - cos(12x)]= (1/2) * (cos2x - cos12x)∫ sin5x * sin7x dx= (1/2)∫ (cos2x - cos12x) dx= (1/2) * (1/2*sin2x - 1/12*sin12x) c= (1/4)sin2x - (1/24)sin12x c= (1/24) * (6sin2x - sin12x) c(4):cos²(a wt) = (1/2) * {1 cos[2(a wt)]}∫ cos²(a wt) dt= (1/2)∫ {1 cos[2(a wt)] dt= (1/2)∫ dt (1/2)∫ cos(2a 2wt) dt= 1/2 * t (1/2) * 1/(2w) * ∫ cos(2a 2wt) d(2a 2wt)= t/2 1/(4w) * sin(2a 2wt) c。
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