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定积分∫(0到π/4)(cosx)^4=
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定积分∫(0到π/4)(cosx)^4=
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答案和解析
∫【0→π/4】(cosx)^4dx
=∫【0→π/4】[(cos2x+1)/2]²dx
=∫【0→π/4】(cos²2x+2cos2x+1)/4 dx
=1/4 ∫【0→π/4】[(cos4x+1)/2+2cos2x+1]dx
=1/4 ∫【0→π/4】[(cos4x)/2+2cos2x+3/2]dx
=【0→π/4】1/4 [(sin4x)/8+sin2x+3x/2]
=1/4[(sinπ)/8+sin(π/2)+3π/8-0]
=1/4+3π/32
=∫【0→π/4】[(cos2x+1)/2]²dx
=∫【0→π/4】(cos²2x+2cos2x+1)/4 dx
=1/4 ∫【0→π/4】[(cos4x+1)/2+2cos2x+1]dx
=1/4 ∫【0→π/4】[(cos4x)/2+2cos2x+3/2]dx
=【0→π/4】1/4 [(sin4x)/8+sin2x+3x/2]
=1/4[(sinπ)/8+sin(π/2)+3π/8-0]
=1/4+3π/32
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