定积分∫(上限π⼀3,下限π⼀4)x⼀(sin^2x)dx

2025-03-10 13:16:40

推荐回答（1个）

回答1：

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