求 （1）∫sin2xcos3x dx （2）∫cosx(cosx/2) dx（...
求 （1）∫sin2xcos3x dx （2）∫cosx(cosx/2) dx（3）∫sin5xsin7x dx （4）∫cos²（wt a）dt
(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。
答：∫sec x dx=∫cosx dx/cosx.cosx ==∫dsinx/(1-sinx)1+sinx) =(In|1+sinx|-In|1-sinx|)/2...详情>>
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