定积分∫(-π/2,π/2)(cos^4x+sin^3x)dx=

定积分∫(-π/2,π/2)(cos^4x+sin^3x)dx=
定积分∫(-π/2,π/2)(cos^4x+sin^3x)dx=

定积分∫(-π/2,π/2)(cos^4x+sin^3x)dx=
∵(cosx)^4是偶函数,(sinx)^3是奇函数
∴∫(cosx)^4dx=2∫(cosx)^4dx
∫(sinx)^3dx=0
故 ∫((cosx)^4+(sinx)^3)dx
=∫(cosx)^4dx+∫(sinx)^3dx
=2∫(cosx)^4
=(1/2)∫[3/2+2cos(2x)+cos(4x)/2]dx (应用倍角公式)
=(1/2)[3x/2+sin(2x)+sin(4x)/8]│
=(1/2)(3π/4)
=3π/8.

既然你知道(sinx)^3是奇函数，那么就等于积分第一项，用两次倍角公式不就行了？
(cosx)^2=1/2*cos2x+1/2

∫(-π/2->π/2)(cosx)^4+(sinx)^3 ]dx
=∫(-π/2->π/2)(sinx)^3 dx
=2∫(0->π/2)(sinx)^3 dx
=-2∫(0->π/2)[1-(cosx)^2 ] dcosx
=-2[ cosx - (cosx)^3/3](0->π/2)
= 2( 1 -1/3)
= 4/3不对 sinX是奇...

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∫(-π/2->π/2)(cosx)^4+(sinx)^3 ]dx
=∫(-π/2->π/2)(sinx)^3 dx
=2∫(0->π/2)(sinx)^3 dx
=-2∫(0->π/2)[1-(cosx)^2 ] dcosx
=-2[ cosx - (cosx)^3/3](0->π/2)
= 2( 1 -1/3)
= 4/3

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