定积分∫(上限π/3,下限π/4)x/(sin^2x)dx

jooops 1年前 已收到1个回答 举报

P鲁罗舒贝 幼苗

共回答了22个问题采纳率:95.5% 举报

原式=∫x*csc^2x dx(下限π/4,上限π/3)
=-(1/2)*∫xd(cot2x)(下限π/4,上限π/3)
=-(1/2)*xcot2x+(1/2)*∫cot2xdx(下限π/4,上限π/3)
=-(1/2)*(π/3)*cot(2π/3)+(1/4)*∫(cos2x/sin2x)d(2x)(下限π/4,上限π/3)
=√3π/18+(1/4)*∫d(sin2x)/sin2x(下限π/4,上限π/3)
=√3π/18+(1/4)*ln|sin2x|(下限π/4,上限π/3)
=√3π/18+(1/4)*ln(√3/2)=√3π/18+(ln3)/8-(ln2)/4

1年前

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