Question

高数，第四题怎么做？

Answer 1

如图

Answer 2

1/[sin2x+2sinx]
=1/[2sinxcosx+2sinx]
=1/[2sinx(1+cosx)]（上下都乘以sinx）
=sinx/[2sinx*sinx*(1+cosx)]
所以
∫dx/sin2x+2sinx
=1/2∫sinx/[(1-(cosx)^2)(1+cosx)]dx
=-1/2∫1/[(1-(cosx)^2)(1+cosx)]dcosx(凑微分法,记cosx=t)
=-1/2∫1/[(1-t^2)(1+t)]dt
=-1/2{-1/4*ln(t-1)-1/2*1/(1+t)+1/4*ln(1+t)}+C
=1/8*(ln(cosx-1)+ln(cosx-1)*cosx+2-ln(1+cosx)-ln(1+cosx)*cosx)/(1+cosx)+C

Answer 3

(4) I = ∫dx/(sin2x+2sinx) = ∫dx/[2sinx(1+cosx)] = (1/2)∫(1-cosx)dx/(sinx)^3
= (1/2)∫(cscx)^3dx - (1/2)∫dsinx/(sinx)^3
= (1/2)∫(cscx)^3dx + (1/4)/(sinx)^2
其中 I1= ∫(cscx)^3dx = -∫cscxdcotx
= -cscxcotx - ∫cotx cscxcotxdx
= -cscxcotx - ∫cscx(cotx)^2dx
= -cscxcotx - I + ∫cscxdx
= -cscxcotx - I1 + ln|cscx-cotx|
则 I1 = -(1/2)cscxcotx + (1/2)ln|cscx-cotx|
I = -(1/4)cscxcotx + (1/4)ln|cscx-cotx| + (1/4)(cscx)^2 + C