∫sinx/1+cos^2xdx
sinx/cos^2xdx的不定积分 -
∫sin/(cosx)^2dx =∫1/(cosx)^2d(-cosx)=-∫1/(cosx)^2dcosx =1/cosx+C
=-∫-sinx/cos^2xdx =-∫1/cos^2xd(cosx)=1/cosx+C
∫sin/(cosx)^2dx =∫1/(cosx)^2d(-cosx)=-∫1/(cosx)^2dcosx =1/cosx+C
=-∫-sinx/cos^2xdx =-∫1/cos^2xd(cosx)=1/cosx+C
∫sin/(cosx)^2dx =∫1/(cosx)^2d(-cosx)=-∫1/(cosx)^2dcosx =1/cosx+C
解：原式=∫dx/[sinx(1-2sin²x)=∫[(1/sinx)+(2sinx)/(1-2sin²x)]dx =∫cscxdx+2∫[(sinx)/(cos²x-sin²x)]dx=ln(cscx-cotx)+2∫[(sinx)/(cosx+sinx)(cosx-sinx)]dx =ln(cscx-cotx)+∫[1/(cosx-sinx)-1/(cosx+sinx)]dx =ln(cscx-cotx)+∫是什么。
不定积分∫sinxcos^2xdx分部积分法 -
∫sinxcos^2xdx=∫sinxcosxdsinx=1/2∫cosxdsin^2x =1/2cosxsin^2x+1/2∫sin^3x=1/2cosxsin^2x+1/2∫sinx(1-cos^2x)dx =1/2cosxsin^2x-1/2cosx-1/2∫sinxcos^2xdx 故：∫sinxcos^2xdx=2/3[1/2cosxsin^2x-1/2cosx]+C 有帮助请点赞。
let y= cosx dy = -sinx dx x=0, y= 1 x=π , y=-1 ∫(0->π)sinx/[1+(cosx)^2]dx =∫(-1->1) dy/(1+y^2)=[arctany]|(-1->1)=π/2
∫sinxcos^2xdx= -
∫sinxcos^2xdx=- ∫cos^2xdcosx= 1/3cos^3x
∫cscxdx =∫1/sinxdx =∫sinx/sin^2xdx =-∫1/(1-cos^2x)dcosx =-1/2∫[1/(1-cosx)+1/(1+cosx)]dcosx =1/2ln(1-cosx)-1/2ln(1+cosx)+C
∫1/sinxdx=∫sinx/1-cos^2xdx=-∫1/1-cos^2xdcosx=? 这道 -
∫1/sinxdx=∫sinx/sin^2xdx =-∫dcosx/(1-cos^2x)=-∫dt/(1-t^2)[令t=cosx]=-1/2∫(1/(t+1)-1/(t-1))dt =-1/2(ln|t+1|-ln|t-1|)+C =-1/2ln|(cosx+1)/(cosx-1)|+C
∫sinxcos^2xdx=∫sinxcosxdsinx=1/2∫cosxdsin^2x =1/2cosxsin^2x+1/2∫sin^3x=1/2cosxsin^2x+1/2∫sinx(1-cos^2x)dx =1/2cosxsin^2x-1/2cosx-1/2∫sinxcos^2xdx 故：∫sinxcos^2xdx=2/3[1/2cosxsin^2x-1/2cosx]+C 有帮助请点赞。