∫ x^3sin^2x dx
= ∫ x^3 * (1 - cos(2x))/2 dx
= (1/8)x^4 - (1/2)K
where K = ∫ x^3cos(2x) dx
令f = x^3 and g = cos(2x),以下分别对f求导,对g求积分
f = x^3 g = cos2x
f' = 3x^2 g(1) = (1/2)sin(2x)
f'' = 6x g(2) = (- 1/4)cos(2x)
f''' = 6 g(3) = (- 1/8)sin(2x)
f'''' = 0 g(4) = (1/16)cos(2x)、这个是分部积分法的速解法
于是∫ x^3cos(2x) dx
= [f * g(1)] - [f' * g(2)] + [f'' * g(3)] - [f''' * g(4)]
= [x^3 * (1/2)sin(2x)] - [3x^2 * (- 1/4)cos(2x)] + [6x * (- 1/8)sin(2x)] - [6 * (1/16)cos(2x)]
= (1/2)x^3sin(2x) + (3/4)x^2cos(2x) - (3/4)xsin(2x) - (3/8)cos(2x)
原式 = (1/8)x^4 - (1/4)x^3sin(2x) - (3/8)x^2cos(2x) + (3/8)xsin(2x) + (3/16)cos(2x) + C