题目

求 （1）∫sin2xcos3x dx （2）∫cosx(cosx/2) dx（3）∫sin5xsin7x dx （4）∫cos²（wt+a）dt

提问时间：2021-04-12

答案

(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