Question

求定积分谢谢

Answer 1

∫[π/4：π/3](x/sin²x)dx
=∫[π/4：π/3](x·csc²x)dx
=-∫[π/4：π/3]xd(cotx)
=-x·cotx|[π/4：π/3]+∫[π/4：π/3]cotxdx
=-[(π/3)·cot(π/3)-(π/4)·cot(π/4)] +∫[π/4：π/3](cosx/sinx)dx
=-[(π/3)·(√3/3)-(π/4)·1] +∫[π/4：π/3](1/sinx)d(sinx)
=π/4 - √3π/9 +ln|sinx||[π/4：π/3]
=π/4 - √3π/9+ln(π/3)-ln(π/4)
=π/4 - √3π/9+ln4-ln3

追问

可是答案是这样

追答

答案错了。
π/4 - √3π/9+ln4-ln3=(1/36)[(9-4√3)π+72ln(2/3)]

追问

代入去不应该是lnsinxpi/3吗

追答

已经检查了，确定无误。答案肯定是把加号错搞成减号了。
本题运用分部积分法解题。

追问

Lnpi/3怎得出来的

追答

推荐答案之后，就无法修改了。
已经给你发私信。
再写一下：
是我搞错了。书上的答案无误。重新写一下：
∫[π/4：π/3](x/sin²x)dx
=∫[π/4：π/3](x·csc²x)dx
=-∫[π/4：π/3]xd(cotx)
=-x·cotx|[π/4：π/3]+∫[π/4：π/3]cotxdx
=-[(π/3)·cot(π/3)-(π/4)·cot(π/4)] +∫[π/4：π/3](cosx/sinx)dx
=-[(π/3)·(√3/3)-(π/4)·1] +∫[π/4：π/3](1/sinx)d(sinx)
=π/4 - √3π/9 +ln|sinx||[π/4：π/3]
=π/4 - √3π/9+ln(sinπ/3)-ln(sinπ/4)
=π/4 - √3π/9+ln(√3/2)-ln(√2/2)
=π/4 - √3π/9+½ln(3/2)
=(1/36)[(9-4√3)π+18ln(3/2)]