搜搜考试网求定积分,
来源：学生作业帮 编辑：搜搜考试网作业帮 分类：数学作业 时间：2024/05/04 07:00:07
求定积分,
原积分=
∫dx/[(x-cosα)^2+(sinα)^2]=(1/sinα)∫d[/(x-cosα)/sinα] / {1+[(x-cosα)/sinα]^2}
=(1/sinα) arctan[(x-cosα)/sinα] |(-1->1)
=(1/sinα) {arctan[(1-cosα)/sinα]-arctan[(-1-cosα)/sinα]}
=(1/sinα) {arctan[tan(α/2)]-arctan[tan(π/2+α/2)]}
=(1/sinα)[α/2-(π/2+α/2)]
=-π/(2sinα)
∫dx/[(x-cosα)^2+(sinα)^2]=(1/sinα)∫d[/(x-cosα)/sinα] / {1+[(x-cosα)/sinα]^2}
=(1/sinα) arctan[(x-cosα)/sinα] |(-1->1)
=(1/sinα) {arctan[(1-cosα)/sinα]-arctan[(-1-cosα)/sinα]}
=(1/sinα) {arctan[tan(α/2)]-arctan[tan(π/2+α/2)]}
=(1/sinα)[α/2-(π/2+α/2)]
=-π/(2sinα)