积分[cosx/(sin^2x-6sinx+12)]dx

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积分[cosx/(sin^2x-6sinx+12)]dx

积分[cosx/(sin^2x-6sinx+12)]dx
积分[cosx/(sin^2x-6sinx+12)]dx

积分[cosx/(sin^2x-6sinx+12)]dx
令a=sinx
则原式=∫dsinx/(sin²a-6sina+12)
=∫da/[(a-3)²+3]
=1/3*∫da/[(a-3)²/3+1]
=1/3*√3*∫d(a/√3-√3)/[(a/√3-√3)²+1]
=√3/3*arctan(a/√3-√3)+C
=√3/3*arctan(sinx/√3-√3)+C

sin^2x这是什么意思

∫ {cosx/[ (sinx)^2-6sinx+12 ]} dx
(sinx)^2-6sinx+12 = (sinx - 3)^2 + 3
let
sinx -3 = √3tany
cosxdx = √3 (secy)^2 dy
∫ {cosx/[ (sinx)^2-6sinx+12 ]} dx
= (1/√3)∫ dy
=(1/√3)y + C
=(1/√3)arctan{ (sinx -3)/√3 } + C