微积分求解不定式sin,cos∫ sin^3(2x)cos^3(2x)dx 要怎么算呢?如果u=cos2x,那么du=2sin2xdx,du里面有2,不知道要怎么代进去.
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![微积分求解不定式sin,cos∫ sin^3(2x)cos^3(2x)dx 要怎么算呢?如果u=cos2x,那么du=2sin2xdx,du里面有2,不知道要怎么代进去.](/uploads/image/z/12875791-31-1.jpg?t=%E5%BE%AE%E7%A7%AF%E5%88%86%E6%B1%82%E8%A7%A3%E4%B8%8D%E5%AE%9A%E5%BC%8Fsin%2Ccos%E2%88%AB+sin%5E3%282x%29cos%5E3%282x%29dx+%E8%A6%81%E6%80%8E%E4%B9%88%E7%AE%97%E5%91%A2%3F%E5%A6%82%E6%9E%9Cu%3Dcos2x%2C%E9%82%A3%E4%B9%88du%3D2sin2xdx%2Cdu%E9%87%8C%E9%9D%A2%E6%9C%892%2C%E4%B8%8D%E7%9F%A5%E9%81%93%E8%A6%81%E6%80%8E%E4%B9%88%E4%BB%A3%E8%BF%9B%E5%8E%BB.)
微积分求解不定式sin,cos∫ sin^3(2x)cos^3(2x)dx 要怎么算呢?如果u=cos2x,那么du=2sin2xdx,du里面有2,不知道要怎么代进去.
微积分求解不定式sin,cos
∫ sin^3(2x)cos^3(2x)dx 要怎么算呢?如果u=cos2x,那么du=2sin2xdx,du里面有2,不知道要怎么代进去.
微积分求解不定式sin,cos∫ sin^3(2x)cos^3(2x)dx 要怎么算呢?如果u=cos2x,那么du=2sin2xdx,du里面有2,不知道要怎么代进去.
求不定积分∫ sin³(2x)cos³(2x)dx
原式=∫(sin2xcos2x)³dx
=∫[(1/2)sin4x]³dx
=(1/8)∫(sin³4x)dx ] [用公式降幂:sin³α=(1/4)(3sinα-sin3α)],此处α=4x]
=(1/8)∫(1/4)(3sin4x-sin12x)dx
=(1/32)[3∫sin4xdx-∫sin12xdx]
=(1/32)[(3/4)∫sin4xd(4x)-(1/12)∫sin12xd(12x)]
=(1/32)[-(3/4)cos4x+(1/12)cos12x]+C
=(1/384)(-9cos4x+cos12x)+C
要先简化原式
sin³2xcos³2x = (1/8)(2sin2xcos2x)³ = (1/8)sin³4x
= (1/8)(sin²4x)sin4x = (1/8)(1 - cos²4x)sin4x
= (1/8)[1 - (1 + cos8x)/2]sin4x
= (1/16)(1- cos8x)si...
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要先简化原式
sin³2xcos³2x = (1/8)(2sin2xcos2x)³ = (1/8)sin³4x
= (1/8)(sin²4x)sin4x = (1/8)(1 - cos²4x)sin4x
= (1/8)[1 - (1 + cos8x)/2]sin4x
= (1/16)(1- cos8x)sin4x
= (1/16)sin4x - (1/16)cos8xsin4x
= (1/16)sin4x - (1/32)(sin12x - sin4x)
= (3/32)sin4x - (1/32)sin12x
积分= -(3/128)cos4x + (1/384)cos12x + C
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