∫cosx／﹙2sinx-cosx﹚dx怎样求解

观察d(2sinx - cosx) = (2cosx + sinx) dx
分子可变为cosx = A(2cosx + sinx) + B(2sinx - cosx) + C的形式
cosx = (2A - B)cosx + (A + 2B)sinx + C
C = 0
A + 2B = 0、A = - 2B
2A - B = 1、2(- 2B) - B = 1、- 5B = 1、B = - 1/5
A = - 2(- 1/5) = 2/5
∴
∫ cosx/(2sinx - cosx) dx
= (2/5)∫ (2cosx + sinx)/(2sinx - cosx) dx + (- 1/5)∫ (2sinx - cosx)/(2sinx - cosx) dx
= (2/5)∫ d(2sinx - cosx)/(2sinx - cosx) - (1/5)∫ dx
= (2/5)ln|2sinx - cosx| - x/5 + C

2sinx-cosx=√5sin(x-a) (cosa=2/√5,sina=-1/√5)
cosx=cos(x-a+a)=cos(x-a)cosa+sin(x-a)sina
∫cosx／﹙2sinx-cosx﹚dx
=∫[cos(x-a)cosa+sin(x-a)sina]/√5sin(x-a) *dx
=(1/√5)cosaln|sin(x-a)|+(1/√5)xsina+C1
=(1/5)[2ln|(2sinx-cosx)/√5|-x]+C1
=(1/5)[2ln|2sinx-cosx|-x]+C

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