Question

积分区间是0到二分之一π，求（sinx）^4（cosx）^4dx的定积分？

改成积分区间是0到二分之一π，求（sinx）^4（cosx）^2dx的定积分？

Answer 1

解：∵(sinx)^4(cosx)^4=[sin(2x)/2]^4
=[sin²(2x)]²/2^4
=[(1-cos(4x))/2]²/2^4
=[1-2cos(4x)+cos²(4x)]/2^6
=[1-2cos(4x)+(1+cos(8x))/2]/2^6
=[3-4cos(4x)+cos(8x)]/2^7
∴∫<0,π/2>(sinx)^4(cosx)^4dx=(1/2^7)∫<0,π/2>[3-4cos(4x)+cos(8x)]dx
=(1/2^7)[3x-sin(4x)+sin(8x)/8]│<0,π/2>
=(1/2^7)(3π/2)
=3π/2^8
=3π/256。