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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
人气:444 ℃ 时间:2020-02-03 07:48:18
解答
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
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