不定积分①∫(sin2x)⼀(sin눀x)dx ②∫sin대xcos눀xdx

2025-12-16 18:28:31
推荐回答(1个)
回答1:

① ∫ (sin2x)/(sin²x) dx

= ∫ (2sinxcosx)/(sin²x) dx

= 2∫ cosx/sinx dx

= 2∫ (1/sinx) d(sinx)

= 2ln|sinx| + C

②∫ sin³xcos²x dx

= ∫ sin²xcos²x d(-cosx)

= -∫ (1 - cos²x)cos²x d(cosx)

= ∫ (cos⁴x - cos²x) d(cosx)

= (1/5)cos⁵x - (1/3)cos³x + C