题目

∫cos^2(ωt+ψ）sin(ωt+ψ）dt 这个的不定积分怎么求?谢谢啦

提问时间：2020-10-31

答案

设ωt+ψ = θ => ωdt = dθ => dt = (1/ω)dθ
∫cos²(ωt+ψ)sin(ωt+ψ) dt = ∫cos²θsinθ * (1/ω)dθ
= (1/ω)∫cos²θ d(-cosθ)
= (-1/ω)*(1/3)cos³θ + C
= [-1/(3ω)]cos³(ωt+ψ) + C