求不定积分:1/(sinx)^3不查表怎么去求它的原函数啊
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![求不定积分:1/(sinx)^3不查表怎么去求它的原函数啊](/uploads/image/z/10164164-68-4.jpg?t=%E6%B1%82%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%EF%BC%9A1%2F%EF%BC%88sinx%EF%BC%89%5E3%E4%B8%8D%E6%9F%A5%E8%A1%A8%E6%80%8E%E4%B9%88%E5%8E%BB%E6%B1%82%E5%AE%83%E7%9A%84%E5%8E%9F%E5%87%BD%E6%95%B0%E5%95%8A)
求不定积分:1/(sinx)^3不查表怎么去求它的原函数啊
求不定积分:1/(sinx)^3
不查表怎么去求它的原函数啊
求不定积分:1/(sinx)^3不查表怎么去求它的原函数啊
∫1/(sinx)^3dx
=∫1/(sinx(1-(cosx)^2))dx
=(1/2)∫(1/sinx)[1/(1+cosx)+1/(1-cosx)]dx
=(1/2)∫sinx/(sinxsinx(1+cosx))dx+(1/2)∫sinx/(sinxsinx(1-cosx))dx
=(1/2)∫sinx/(1-(cosx)^2)(1+cosx))dx+(1/2)∫sinx/((1-(cosx)^2)(1-cosx))dx
=-(1/2)∫1/((1-(cosx)^2)(1+cosx))d(cosx)-(1/2)∫1/((1-(cosx)^2)(1-cosx))d(cosx)
=(-1/2)∫(1/2)(1/(1+cosx)+1/(1-cosx))(1/(1+cosx))dcosx
+(-1/2)∫(1/2)(1/(1+cosx)+1/(1-cosx))(1/(1-cosx))dcosx
=(1/4)(1/(1+cosx))-1/8ln(1+cosx)+(1/8)ln(1-cosx)-(1/4)1/(1-cosx)
-1/8ln(1+cosx)+(1/8)ln(1-cosx)
=-cosx/(sinx)^2/2+ln((1-cosx)/sinx)/2+C
∫1/(sinx)^3dx=∫sinx/(sinx)^4dx
=∫sinx/(1-cosx^2)^2dx
=-∫1/(1-cosx^2)^2dcosx
1/(1-u^2)^2=A/(1-u)^2+B/(1+u)^2+C/(1-u)+D/(1+u)
A=1/4,B=1/4,C=1/4,D=1/4;
∫1/(1-u^2)^2du=1/4[(/1-u)-(1/1+u)-ln|1-u|+ln|1+u|]+C( |u|<1)
将COSx代入即得