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

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

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

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

sin(ωt+ψ)dt=-[dcos(ωt+ψ)]/ω
所以∫cos^2(ωt+ψ)sin(ωt+ψ)dt =-cos^3(ωt+ψ)/(3ω) +C

t^1/2 dt怎么求啊? ∫ sin(t^1/2) / t^1/2 dt =∫ sin([t^(-1/2) dt] = 2 ∫ sin(t^1/2) * dt^(1/2) = -2 cos

∫cos^2(ωt+ψ)sin(ωt+ψ)dt
=(-1/ω)∫cos^2(ωt+ψ)dcos(ωt+ψ)
=(-1/(3ω)) *cos^3(ωt+ψ)

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