第1个回答 2019-03-06
原式=∫（0，π/2）-cos^(5)x d(cosx)
=-cos^(6)x/6|（0，π/2）
=-cos^(6)(π/2)/6+cos^6(0)/6
=0+1/6
=1/6
=-cos^(6)x/6|（0，π/2）
=-cos^(6)(π/2)/6+cos^6(0)/6
=0+1/6
=1/6
第2个回答 2019-03-06
∫(0->π/2) (cosx)^5.sinx dx
=-∫(0->π/2) (cosx)^5 dcosx
=-(1/6)[ (cosx)^6 ]|(0->π/2)
=1/6
=-∫(0->π/2) (cosx)^5 dcosx
=-(1/6)[ (cosx)^6 ]|(0->π/2)
=1/6
第3个回答 2019-03-06
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