∫x^2sin(4x)dx 怎么算啊
人气:486 ℃ 时间:2020-09-05 13:59:38
解答
∫x^2sin(4x)dx =1/4∫x^2sin(4x)d4x =-1/4∫x^2dcos(4x) =-1/4*x^2cos4x+1/4∫cos4xdx^2 =-1/4*x^2cos4x+1/2∫xcos4xdx =-1/4*x^2cos4x+1/8∫xcos4xd4x =-1/4*x^2cos4x+1/8∫xdsin4x =-1/4*x^2cos4x+1/8*xsin4x-1/8...
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