求不定积分∫(1/cosx)dx书里解题是=∫dx/sin(x+π/2)=ln|tan(x/2+π/4)|+c我搞不清这题的公式是怎么转变的?cosx怎么变成了sin(x+π/2),sin(x+π/2)又怎么积分成最后的答案lntan?
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![求不定积分∫(1/cosx)dx书里解题是=∫dx/sin(x+π/2)=ln|tan(x/2+π/4)|+c我搞不清这题的公式是怎么转变的?cosx怎么变成了sin(x+π/2),sin(x+π/2)又怎么积分成最后的答案lntan?](/uploads/image/z/5392668-12-8.jpg?t=%E6%B1%82%E4%B8%8D%E5%AE%9A%E7%A7%AF%E5%88%86%E2%88%AB%EF%BC%881%2Fcosx%EF%BC%89dx%E4%B9%A6%E9%87%8C%E8%A7%A3%E9%A2%98%E6%98%AF%3D%E2%88%ABdx%2Fsin%28x%2B%CF%80%2F2%29%3Dln%7Ctan%28x%2F2%2B%CF%80%2F4%29%7C%2Bc%E6%88%91%E6%90%9E%E4%B8%8D%E6%B8%85%E8%BF%99%E9%A2%98%E7%9A%84%E5%85%AC%E5%BC%8F%E6%98%AF%E6%80%8E%E4%B9%88%E8%BD%AC%E5%8F%98%E7%9A%84%3Fcosx%E6%80%8E%E4%B9%88%E5%8F%98%E6%88%90%E4%BA%86sin%28x%2B%CF%80%2F2%29%2Csin%28x%2B%CF%80%2F2%29%E5%8F%88%E6%80%8E%E4%B9%88%E7%A7%AF%E5%88%86%E6%88%90%E6%9C%80%E5%90%8E%E7%9A%84%E7%AD%94%E6%A1%88lntan%3F)
求不定积分∫(1/cosx)dx书里解题是=∫dx/sin(x+π/2)=ln|tan(x/2+π/4)|+c我搞不清这题的公式是怎么转变的?cosx怎么变成了sin(x+π/2),sin(x+π/2)又怎么积分成最后的答案lntan?
求不定积分∫(1/cosx)dx
书里解题是=∫dx/sin(x+π/2)=ln|tan(x/2+π/4)|+c
我搞不清这题的公式是怎么转变的?
cosx怎么变成了sin(x+π/2),sin(x+π/2)又怎么积分成最后的答案lntan?
求不定积分∫(1/cosx)dx书里解题是=∫dx/sin(x+π/2)=ln|tan(x/2+π/4)|+c我搞不清这题的公式是怎么转变的?cosx怎么变成了sin(x+π/2),sin(x+π/2)又怎么积分成最后的答案lntan?
sin(x+π/2)=sinxcosπ/2+cosxsinπ/2=cosx
∫dx/sin(x+π/2)=∫dx/[2sin(x/2+π/4)cos(x/2+π/4)]
=∫cos(x/2+π/4)dx/[2sin(x/2+π/4)[cos(x/2+π/4)]^2]
=2∫dsin(x/2+π/4)/[2sin(x/2+π/4)[1-[sin(x/2+π/4)]^2]](令sin(x/2+π/4)=t)
=∫dt/[t(1-t^2)]
=∫dt/t+∫t*dt/(1-t^2)
=ln|t|-(1/2)*∫d(1-t^2)/(1-t^2)
=ln|t|-(1/2)*ln|1-t^2|+C
=ln(t/(1-t^2)^(1/2))+C
将sin(x/2+π/4)带入
得
ln|tan(x/2+π/4)|+c
这是因为∫(1/sinx)dx有一个现成的公式可用
∫(1/sinx)dx
=∫dx/[2sin(x/2)cos(x/2)]
=1/2*∫[(secx/2)^2/tan(x/2)]dx
=∫dtan(x/2)/tan(x/2)=ln|tan(x/2)|+C
可以上下乘以cosx
∫(1/cosx)dx=∫cosxdx/(cosx)^2=∫dsinx/(1-sinx*sinx)
这样就简单了~