来自胡云安的问题
∫(0,π/4)√(1-cos2α)dα=
∫(0,π/4)√(1-cos2α)dα=
3回答
2020-11-0900:23
∫(0,π/4)√(1-cos2α)dα=
∫(0,π/4)√(1-cos2α)dα=
先求不定积分:
∫√(1-cos2α)dα
=∫√(2(sinα)^2)dα
(因为:α属于(0,π/4)
所以:sinα>0)
=∫√2*sinαdα
=-√2*cosα+C(C是任意常数)
再求定积分:
∫(0,π/4)√(1-cos2α)dα
=-√2[cos0-cos(π/4)]
=1-√2
你最后减反了吧
呵呵不好意思积分上限应该是:π/4下限才是0所以应该是:-√2[cos(π/4)-cos0]=√2-1