Question

如图，请问这个不定积分怎么求解呢？

Answer 1

I = ∫x|cosxsinx|dx = (1/2)∫x|sin2x|dx
当 2kπ ≤ 2x ≤ 2kπ+π， 即 kπ ≤ x ≤ kπ+π/2 时
I = (1/2)∫xsin2xdx = - (1/4)∫xdcos2x
= - (1/4)xcos2x + (1/4)∫cos2xdx = - (1/4)xcos2x + (1/8)sin2x + C
当 2kπ-π ≤ 2x ≤ 2kπ， 即 kπ-π/2 ≤ x ≤ kπ 时
I = - (1/2)∫xsin2xdx = (1/4)∫xdcos2x
= (1/4)xcos2x - (1/4)∫cos2xdx = (1/4)xcos2x - (1/8)sin2x + C。

Answer 2

I = ∫x|cosxsinx|dx = (1/2)∫x|sin2x|dx
当 2kπ ≤ 2x ≤ 2kπ+π， 即 kπ ≤ x ≤ kπ+π/2 时
I = (1/2)∫xsin2xdx = - (1/4)∫xdcos2x
= - (1/4)xcos2x + (1/4)∫cos2xdx = - (1/4)xcos2x + (1/8)sin2x + C
当 2kπ-π ≤ 2x ≤ 2kπ， 即 kπ-π/2 ≤ x ≤ kπ 时
I = - (1/2)∫xsin2xdx = (1/4)∫xdcos2x
= (1/4)xcos2x - (1/4)∫cos2xdx = (1/4)xcos2x - (1/8)sin2x + C

Answer 3