(x+cosx^2)sinx^4dx 在-π/2 到 π/2 上的定积分
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(x+cosx^2)sinx^4dx 在-π/2 到 π/2 上的定积分
(x+cosx^2)sinx^4dx 在-π/2 到 π/2 上的定积分
(x+cosx^2)sinx^4dx 在-π/2 到 π/2 上的定积分
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∫(-π/2->π/2)(x+(cosx)^2)(sinx)^4 dx
= ∫(-π/2->π/2)x(sinx)^4 dx + ∫(-π/2->π/2)(cosx)^2(sinx)^4 dx
=∫(-π/2->π/2)x(sinx)^4 dx
=(1/4) ∫(-π/2->π/2) x ( 1-cos2x)^2 dx
=(1/4) ∫(-π/2->π/2) x (...
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∫(-π/2->π/2)(x+(cosx)^2)(sinx)^4 dx
= ∫(-π/2->π/2)x(sinx)^4 dx + ∫(-π/2->π/2)(cosx)^2(sinx)^4 dx
=∫(-π/2->π/2)x(sinx)^4 dx
=(1/4) ∫(-π/2->π/2) x ( 1-cos2x)^2 dx
=(1/4) ∫(-π/2->π/2) x ( 1-2cos2x+ (cos2x)^2) dx
= (1/8)∫(-π/2->π/2) x ( 3-6cos2x+ cos4x ) dx
= (1/8) [ 3π^2/4 - ∫(-π/2->π/2) 6xcos2x dx + ∫(-π/2->π/2) xcos4x dx ]
∫(-π/2->π/2) 6xcos2x dx
= 6∫(0->π/2) x dsin2x
= 6{ [xsin2x](0->π/2) - ∫(0->π/2) sin2xdx }
= 3 ( [cos2x](0->π/2) )
= -6
∫(-π/2->π/2) xcos4x dx
=(1/2)∫(0->π/2) xdsin4x
=-(1/2) ∫(0->π/2) sin4x dx
= (1/8)[cos4x](0->π/2)
=0
∫(-π/2->π/2)(x+(cosx)^2)(sinx)^4 dx
= (1/8) [ 3π^2/4 - ∫(-π/2->π/2) 6xcos2x dx + ∫(-π/2->π/2) xcos4x dx ]
=(1/8)(3π^2/4+6)
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