indefinite integration

a) ∫sin3xsinx dx
= 1/2 ∫(sin4x + sin2x) dx
= 1/2[(-1/4)cos4x - (1/2)cos2x] + C
= (-1/8)cos4x - (1/4)cos2x + C

b) (cos5x - cosx) / cosx
= [-2sin(5x+x)/2 * sin(5x-x)/2] / cosx
= -2sin3xsin2x/cosx
= -2sin3x(2sinxcosx)/cosx
= -4sin3xsinx

c) (cos5x - cosx) / cosx = -4sin3xsinx
cos5x/cosx - 1 = -4sin3xsinx
cos5x/cosx = 1 - 4sin3xsinx
∫(cos5x)/cosx dx
= ∫(1 - 4sin3xsinx) dx
= x - 4[(-1/8)cos4x - (1/4)cos2x] + C
= x + (1/2)cos4x - cos2x + C


2011-03-26 17:48:32 補充：
(c) ans should be
x + (1/2)cos4x + cos2x + C

2011-03-27 11:40:11 補充：
sorry.. 記錯sum and product rule
(a) ∫sin3xsinx dx
= 1/2∫(cos2x - cos4x) dx
= 1/2[(1/2)sin2x - (1/4)sin4x] + C
= (1/4)sin2x - (1/8)sin4x + C

(c) ∫(cos5x)/cosx dx
= ∫(1 - 4sin3xsinx) dx
= x - 4[(1/4)sin2x - (1/8)sin4x] + C
= x - sin2x + (1/2)sin4x + C