帮忙看下这道二重积分计算题目 急需谢谢～
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∫<0,π/2>dx∫<cosx，1>(y^4)dy
=(1/5)∫<0,π/2>[1-(cosx)^5]dx
=(1/5)[∫<0,π/2>dx-∫<0,π/2>(1-sin²x)²d(sinx)]
=(π/10)-(1/5)∫<0,π/2>[1-2sin²x+(sinx)^4]d(sinx)
=(π/10)-(1/5)[sinx-(2/3)sin³x+(1/5)(sinx)^5]<0,π/2>
=(π/10)-(1/5)[1-(2/3)+(1/5)]=(π/10)-(1/5)×(8/15)=(π/10)-(8/75);
=(1/5)∫<0,π/2>[1-(cosx)^5]dx
=(1/5)[∫<0,π/2>dx-∫<0,π/2>(1-sin²x)²d(sinx)]
=(π/10)-(1/5)∫<0,π/2>[1-2sin²x+(sinx)^4]d(sinx)
=(π/10)-(1/5)[sinx-(2/3)sin³x+(1/5)(sinx)^5]<0,π/2>
=(π/10)-(1/5)[1-(2/3)+(1/5)]=(π/10)-(1/5)×(8/15)=(π/10)-(8/75);
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