Question

求助这道题怎么做，高数

Answer 1

(1) [π/6,π/3] ∫cos²x/[x*(π-2x)] dx
= [π/6,π/3] ∫cos²x*[1/x + 2/(π-2x)] dx
= [π/6,π/3] ∫cos²x/x *dx + [π/6,π/3 ∫cos²x/(π/2-x) *dx
(2) 对积分式第一部分积分变量变为u,u=x
[π/6,π/3] ∫cos²x/x *dx 
= [π/6,π/3] ∫cos²u/u *du
(3) 对于积分式的第二部分作变量代换 令u=π/2-x ==> dx = -du,则
[π/6,π/3] ∫cos²x/(π/2-x) *dx
= [π/3,π/6] ∫cos²(π/2-u)/u *(-du) //注意积分限的变化
= [π/6,π/3] ∫sin²u/u * du 
(4) 两部分合并得到：
原积分式= [π/6,π/3] ∫cos²u/u *du + [π/6,π/3] ∫sin²u/u * du 
= [π/6,π/3] ∫(sin²u+cos²u)/u * du 
= [π/6,π/3]∫1/u * du = ln2