大一高数 广义二重积分 求大佬解答!!
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积分域 D 即 π/4 到 π/2 的无限扇形
I = ∫<π/4, π/2>dt∫<0, +∞>e^[-r(cost+sint)] rdr
其中 ∫<0, +∞>e^[-r(cost+sint)] rdr
= -[1/(cost+sint)]∫<0, +∞>r de^[-r(cost+sint)]
= -[1/(cost+sint)][re^{-r(cost+sint)}]<0, +∞>
+ [1/(cost+sint)]∫<0, +∞>e^[-r(cost+sint)]dr
= 0 - [1/(cost+sint)^2] [e^{-r(cost+sint)}]<0, +∞>
= 1/(cost+sint)^2,
I = ∫<π/4, π/2>dt/(cost+sint)^2 = ∫<π/4, π/2>dt/(1+sin2t)
令 tant = u, 则 sin2t = 2u/(1+u^2), dt = 2du/(1+u^2).
I = ∫<1, +∞>2du/(1+u)^2 = [-2/(1+u)]<1, +∞> = 1
I = ∫<π/4, π/2>dt∫<0, +∞>e^[-r(cost+sint)] rdr
其中 ∫<0, +∞>e^[-r(cost+sint)] rdr
= -[1/(cost+sint)]∫<0, +∞>r de^[-r(cost+sint)]
= -[1/(cost+sint)][re^{-r(cost+sint)}]<0, +∞>
+ [1/(cost+sint)]∫<0, +∞>e^[-r(cost+sint)]dr
= 0 - [1/(cost+sint)^2] [e^{-r(cost+sint)}]<0, +∞>
= 1/(cost+sint)^2,
I = ∫<π/4, π/2>dt/(cost+sint)^2 = ∫<π/4, π/2>dt/(1+sin2t)
令 tant = u, 则 sin2t = 2u/(1+u^2), dt = 2du/(1+u^2).
I = ∫<1, +∞>2du/(1+u)^2 = [-2/(1+u)]<1, +∞> = 1
追问
。。。我这儿写的答案是1/2,有可能是它错了。。我再算一遍。
谢谢啦
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