∫(1/sinx)³dx是多少?
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发布时间:2022-05-14 07:10
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热心网友
时间:2023-10-09 13:45
J = ∫ (1/sinx)³ dx
= ∫ csc³x dx
= ∫ cscx d(-cotx)
= -cscxcotx + ∫ cotx(-cscxcotx)dx
= -cscxcotx - ∫ cot²xcscxdx
= -cscxcotx - ∫ (csc²x-1)cscxdx
= -cscxcotx - J + ∫ cscxdx
2J = -cscxcotx + ∫ cscx(cscx+cotx)/(cscx+cotx) dx
J = -(1/2)cscxcotx + (1/2)∫ (cscxcotx+csc²x)/(cscx+cotx) dx
= -(1/2)cscxcotx + (1/2)∫ -d(cscx+cotx)/(cscx+cotx) dx
= -(1/2)cscxcotx - (1/2)ln| cscx+cotx | + C
热心网友
时间:2023-10-09 13:45
∫1/(sinx)^3dx=-∫1/(sinx)^4dcosx=-∫1/[1-(cosx)^2]^2dcosx
设cosx=y
上式=-∫1/[1-y^2]^2dy=-1/2∫(1/(1-y^2)+1/(1+y^2)]dy=-1/2arctany+ln|(y+1)/(y-1)|+c
∫1/(sinx)^3dx=-1/2arctan(cosx)+2ln|(cosx+1)/(cosx-1)|+c