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∫(0%π/2)xsin2xDx的不定积分的详细过程,求% %.....

能拍题吗 看不懂

1、本题是典型的用分部积分的类型; 积分过程还用到了国内盛行的凑微分方法。 2、具体解答如下,如有疑问,欢迎追问,有问必答,有疑必释。 3、若点击放大,图片将会更加清晰。

如图所示:

能拍题吗 看不懂

今天老师刚讲了,我们月考就考了这道题

∫xsin2xdx =(-1/2)∫xdcos2x =(-1/2)(xcos2x-∫cos2xdx) =(-1/2)(xcos2x-(1/2)sin2x)+C =(1/4)sin2x-(1/2)xcos2x+C 代入上下限 =0+π/4-0 =π/4

∫(0->π/2) xcos2xdx =(1/2)∫(0->π/2) xdsin2x =(1/2)[ x.sin2x]|(0->π/2) -(1/2)∫(0->π/2) sin2x dx =0-(1/2)∫(0->π/2) sin2x dx =(1/4)[ cos2x ]|(0->π/2) =-1/2

∫[0→π/2] 2xcos2x dx =∫[0→π/2] x d(sin2x) =xsin2x - ∫[0→π/2] sin2x dx =xsin2x + (1/2)cos2x |[0→π/2] =-(1/2) - (1/2) =-1 希望可以帮到你,不明白可以追问,如果解决了问题,请点下面的"选为满意回答"按钮,谢谢。

∫cos2xdx =∫cos2xd(1/2)2x =-1/2sin2x+C

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