間違いしかありません.コメントにてご指摘いただければ幸いです(気が付いた点を特に断りなく頻繁に書き直していますのでご注意ください).

cosの逆数(sec)の積分

1cos(θ)dθ=sec(θ)dθsec(θ)=1cos(θ)=1cos(θ)cos(θ)cos(θ)dθ=cos(θ)cos2(θ)dθ=cos(θ)1sin2(θ)dθcos2(θ)=1sin2(θ)=12{cos(θ)1+sin(θ)+cos(θ)1sin(θ)}dθcos(θ)1sin2(θ)=α1+sin(θ)+β1sin(θ)cos(θ)=α(1sin(θ))+β(1+sin(θ))=(α+β)+(αβ)sin(θ){cos(θ)=α+β0=(αβ)sin(θ)sin(θ)θ0,αβ0,α=β.α+β=α+α=2α=cos(θ).α=β=12cos(θ)cos(θ)1sin2(θ)=12cos(θ)1+sin(θ)+12cos(θ)1sin(θ)=12{cos(θ)1+sin(θ)+cos(θ)1sin(θ)}dθ=12{cos(θ)1+sin(θ)dθ+cos(θ)1sin(θ)dθ}=12{cos(θ)u1cos(θ)du+cos(θ)v1cos(θ)dv}u=1+sin(θ),dudθ=cos(θ),dθ=1cos(θ)duv=1sin(θ),dvdθ=cos(θ),dθ=1cos(θ)dv=12(1udu1vdv)=12(ln|u|ln|v|)1xdx=ln|x|+C(C:)=12ln|u||v|logAlogB=logAB=12ln|uv||A||B|=|AB|=12ln|1+sin(θ)1sin(θ)|+CC:=12ln|1+sin(θ)1sin(θ)|+CAlogB=logBA=12ln|{1cos(θ)+tan(θ)}2|+C1+sin(θ)1sin(θ)={1cos(θ)+tan(θ)}2=122ln|1cos(θ)+tan(θ)|+C(AB)1B=A(B1B)=A1=A=ln|1cos(θ)+tan(θ)|+C=ln|sec(θ)+tan(θ)|+C1cos(θ)=sec(θ)
1+sin(θ)1sin(θ)=1+sin(θ)1sin(θ)1+sin(θ)1+sin(θ)={1+sin(θ)}21sin2(θ)(A+B)(AB)=A2B2=1+2sin(θ)+sin2(θ)cos2(θ)1sin2(θ)=cos2(θ),(A+B)2=A2+2AB+B2=1cos2(θ)+2sin(θ)cos2(θ)+sin2(θ)cos2(θ)=1cos2(θ)+21cos(θ)sin(θ)cos(θ)+sin2(θ)cos2(θ)=1cos2(θ)+21cos(θ)tan(θ)+tan2(θ)sin(θ)cos(θ)=tan(θ)={1cos(θ)+tan(θ)}2A2+2AB+B2=(A+B)2

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