Question

关于高等数学带绝对值定积分的题目

∫<π/2,-π/2>x+|sinx|/1+cosxdx

Answer 1

结果是2ln2。

解答见图，点击放大：

Answer 2

绝对值|sinx|在区间[-π/2,π/2]符号变化，分成两段求积分就好了

Answer 3

分区间0到二分之派，负的到0

Answer 4

想为你解答但是上传不了图片
结果=ln2

你可以加我QQ 474842291

Answer 5

这绝对值是全看sinx的图像而取决的。
y = (x + |sinx|)/(1 + cosx)、- π/2 ≤ x ≤ π/2
当- π/2 ≤ x ≤ 0、|sinx| = - sinx、因为在[- π/2，0]的sinx ≤ 0
y = (x - sinx)/(1 + cosx)
当0 ≤ x ≤ π/2、|sinx| = sinx、因为在[0，π/2]的sinx ≥ 0
y = (x + sinx)/(1 + cosx)

∫[- π/2→π/2] (x + |sinx|)/(1 + cosx) dx
= ∫[- π/2→0] (x - sinx)/(1 + cosx) dx + ∫[0→π/2] (x + sinx)/(1 + cosx) dx
= A + B
A = ∫[- π/2→0] (x - sinx)/(1 + cosx) dx
= ∫[- π/2→0] x/(1 + cosx) dx - ∫[- π/2→0] sinx/(1 + cosx) dx
= ∫[- π/2→0] x/(1 + cosx) dx - [xsinx/(1 + cosx)] |[- π/2→0] + ∫[- π/2→0] x/(1 + cosx) dx
= - [xtan(x/2)] + 2∫[- π/2→0] x/[2cos²(x/2)] dx
= (- π/2)tan(- π/4) + 2∫[- π/2→0] x d[tan(x/2)]
= π/2 + 2[xtan(x/2)] |[- π/2→0] - 4∫[- π/2→0] tan(x/2) d(x/2)
= π/2 - 2(- π/2)tan(- π/4) + 4ln[cos(x/2)] |[- π/2→0]
= π/2 - π + 2ln[1] - 4ln[1/√2]
= 2ln[2] - π/2
B = ∫[0→π/2] (x + sinx)/(1 + cosx) dx
= ∫[0→π/2] x/(1 + cosx) dx + ∫[0→π/2] sinx/(1 + cosx) dx
= ∫[0→π/2] x/(1 + cosx) dx + [xsinx/(1 + cosx)] |[0→π/2] - ∫[0→π/2] x/(1 + cosx) dx
= [xtan(x/2)] |[0→π/2]
= (π/2)tan(π/4)
= π/2
原式 = 2ln[2] - π/2 + π/2
= 2ln[2]