Question

不定积分转换？

如图，请给出步骤

Answer 1

简单计算一下即可，答案如图所示

Answer 2

先拆分，再换元得到的。

Answer 3

原式=∫(-π/4,0)dx/(1+sinx)+∫(0,π/4)dx/(1+sinx)。对前一个积分令x=-t、合并即可。

Answer 4

Answer 5

let
u=-x
du =-dx
x=-π/4 , u=π/4/4
x=0, u=0
∫(-π/4->π/4) dx/(1-sinx)
=∫(-π/4->0) dx/(1-sinx) +∫(0->π/4) dx/(1-sinx)
=∫(π/4->0) -du/(1+sinu) +∫(0->π/4) dx/(1-sinx)
=∫(0->π/4) du/(1+sinu) +∫(0->π/4) dx/(1-sinx)
=∫(0->π/4) dx/(1+sinx) +∫(0->π/4) dx/(1-sinx)