定积分∫(上限π/3,下限π/4)x/(sin^2x)dx
人气:235 ℃ 时间:2020-02-05 17:31:29
解答
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
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