求定积分∫xsin^2x dx [-π/2,π/2][-π/2,π/2]是区间,因为我不会打,所以~希望有最简洁的答案
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![求定积分∫xsin^2x dx [-π/2,π/2][-π/2,π/2]是区间,因为我不会打,所以~希望有最简洁的答案](/uploads/image/z/5937015-39-5.jpg?t=%E6%B1%82%E5%AE%9A%E7%A7%AF%E5%88%86%E2%88%ABxsin%5E2x+dx+%5B-%CF%80%2F2%2C%CF%80%2F2%5D%5B-%CF%80%2F2%2C%CF%80%2F2%5D%E6%98%AF%E5%8C%BA%E9%97%B4%2C%E5%9B%A0%E4%B8%BA%E6%88%91%E4%B8%8D%E4%BC%9A%E6%89%93%2C%E6%89%80%E4%BB%A5%7E%E5%B8%8C%E6%9C%9B%E6%9C%89%E6%9C%80%E7%AE%80%E6%B4%81%E7%9A%84%E7%AD%94%E6%A1%88)
求定积分∫xsin^2x dx [-π/2,π/2][-π/2,π/2]是区间,因为我不会打,所以~希望有最简洁的答案
求定积分∫xsin^2x dx [-π/2,π/2]
[-π/2,π/2]是区间,因为我不会打,所以~
希望有最简洁的答案
求定积分∫xsin^2x dx [-π/2,π/2][-π/2,π/2]是区间,因为我不会打,所以~希望有最简洁的答案
解法一的(cos2x+1)/2dx 应该是 (1-cos2x)/2dx 高手犯了个低级错误哦!sin^2x=(1-cos2x)/2
解法一
∫xsin^2x dx
=∫x(cos2x+1)/2dx
=(1/4)∫xcos2xd2x
=(1/4)∫xdsin2x
=(1/4)xsin2x-(1/4)∫sin2xdx
=(1/4)xsin2x-(1/8)∫sin2xd2x
=(1/4)xsin2x+(1/8)cos2x [-π/2,π/2]
=[(1/4)(π...
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解法一
∫xsin^2x dx
=∫x(cos2x+1)/2dx
=(1/4)∫xcos2xd2x
=(1/4)∫xdsin2x
=(1/4)xsin2x-(1/4)∫sin2xdx
=(1/4)xsin2x-(1/8)∫sin2xd2x
=(1/4)xsin2x+(1/8)cos2x [-π/2,π/2]
=[(1/4)(π/2)*sinπ+(1/8)cosπ]-[(1/4)(-π/2)*sin(-π)+(1/8)cos(-π)]
=0
解法二
因为f(x)=xsin^2x是奇函数
且积分限关于原点对称
所以积分=0
这个解法二就是简洁的答案
收起
=1/2∫x(1-cos2x)dx
=1/2∫xdx-1/2∫xcos2xdx
=1/4x^2-1/4∫xd(sin2x)
=1/4一1/4xsin2x十1/8cos2x十C