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cos^2(ωx+ψ)dx
人气:338 ℃ 时间:2020-04-06 11:35:52
解答
∫[cos(ωx+Ψ)]^2dx
=(1/2)∫[1+cos2(ωx+Ψ)]dx
=(1/2)∫dx+[1/(4ω)]∫cos2(ωx+Ψ)d[2(ωx+Ψ)]
=(1/2)∫dx+[1/(4ω)]sin2(ωx+Ψ)+C.
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