∫0到2π 根号下1+sinxsinxdx

2025-03-15 22:08:00
推荐回答(2个)
回答(1):

结果为:4/3

解题过程如下:

原式=√[sinx-(sinx)^3]dx ∫(0,π)

=∫(0,π)√[sinx(cosx)^2]

=∫(0,π/2)cosx√sinxdx-∫(π/2,π)cosx√sinxdx

=∫(0,π/2)√sinxdsinx-∫(π/2,π)√sinxdsinx

=(2/3)(sinx)^(3/2)-(2/3)(sinx)^(3/2)

=(2/3)(sinπ/2)^(3/2)-(2/3)(sin0)^(3/2)-(2/3)(sinπ)^(3/2)+(2/3)(sinπ/2)^(3/2)

=2/3-0-0+2/3

=4/3

扩展资料

求函数积分的方法:

设F(x)是函数f(x)的一个原函数,我们把函数f(x)的所有原函数F(x)+C(C为任意常数)叫做函数f(x)的不定积分,记作,即∫f(x)dx=F(x)+C。

其中∫叫做积分号,f(x)叫做被积函数,x叫做积分变量,f(x)dx叫做被积式,C叫做积分常数,求已知函数不定积分的过程叫做对这个函数进行积分。

积分是微积分学与数学分析里的一个核心概念。通常分为定积分和不定积分两种。直观地说,对于一个给定的实函数f(x),在区间[a,b]上的定积分。

若f(x)在[a,b]上恒为正,可以将定积分理解为在Oxy坐标平面上,由曲线(x,f(x))、直线x=a、x=b以及x轴围成的面积值(一种确定的实数值)。

函数积分公式:

回答(2):

如图

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