Question

∫（0，2/π）cos^4xdx的值，提示cos^4x=（cos^2x）^2？

详情见图

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Answer 1

(cosx)^4=[(1+cos2x)/2]^2
=(1+2cos2x+cos^2x)/4
=[1+2cos2x+(1+cos4x)/2]/4
=(3+4cos2x+cos4x)/8,
所以原式=(1/8)[3x+2sin2x+(1/4)sin4x]|<0,π/2>
=(1/8)(3π/2+2）
=3π/16+1/4.

Answer 2

(cosx)^4=[(1+cos2x)/2]^2
=(1+2cos2x+cos^2x)/4
=[1+2cos2x+(1+cos4x)/2]/4
=(3+4cos2x+cos4x)/8,
所以原式=(1/8)[3x+2sin2x+(1/4)sin4x]|<0,π/2>
=(1/8)(3π/2+2）
=3π/16+1/4.