jm=∫0→π xsin∧mx dx
发布网友
发布时间：2022-05-24 10:13
共4个回答
热心网友
时间：2023-10-10 14:32
J(m)=∫(0->π) x.(sinx)^m dx
let
u=π-x
= -dx
x=0，u=π
x=π，u=0
∫(0->π) x.(sinx)^m dx
=∫(π->0) (π-u) .(sinu)^m (-)
=∫(0->π) (π-x) .(sinx)^m dx
2∫(0->π) x.(sinx)^m dx =π∫(0->π) (sinx)^m dx
∫(0->π) x.(sinx)^m dx =(π/2)∫(0->π) (sinx)^m dx
consider
I(m)
=∫(0->π) (sinx)^m dx
=-∫(0->π) (sinx)^(m-1) dcosx
=-[ (sinx)^(m-1).cosx]|(0->π) +(m-1)∫(0->π)(cosx)^2.(sinx)^(m-2) dx
=(m-1)∫(0->π)(cosx)^2.(sinx)^(m-2) dx
=(m-1)∫(0->π)( 1-(sinx)^2).(sinx)^(m-2) dx
2∫(0->π) (sinx)^m dx =(m-1)∫(0->π) (sinx)^(m-2) dx
I(m) = [(m-1)/2] I(m-2)
I(1)=2
I(0) =π
case 1: m 是偶数
I(m)
= [(m-1)/2] I(m-2)
=[(1.3.5...m)/2^(m/2)] I(0)
=[(1.3.5...m)/2^(m/2)] π
J(m)
=∫(0->π) x.(sinx)^m dx
=(π/2)∫(0->π) (sinx)^m dx
=(π/2) I(m)
=(π/2) . { [(1.3.5...m)/2^(m/2)] π }
case 2 : m 是奇数
I(m)
= [(m-1)/2] I(m-2)
={ [(1.3.5...m)/2^[(m-1)/2] } I(1)
={ [(1.3.5...m)/2^[(m-1)/2] } )∫(0->π) (sinx) dx
={ [(1.3.5...m)/2^[(m-1)/2] } . 2
J(m)
=(π/2) I(m)
=π { [(1.3.5...m)/2^[(m-1)/2] }
扩展资料：
一个函数，可以存在不定积分，而不存在定积分，也可以存在定积分，而没有不定积分。连续函数，一定存在定积分和不定积分。
若在有限区间[a,b]上只有有限个间断点且函数有界，则定积分存在；若有跳跃、可去、无穷间断点，则原函数一定不存在，即不定积分一定不存在。
求函数f(x)的不定积分，就是要求出f(x)的所有的原函数，由原函数的性质可知，只要求出函数f(x)的一个原函数，再加上任意的常数C就得到函数f(x)的不定积分。
如果f(x)在区间I上有原函数，即有一个函数F(x)使对任意x∈I，都有F'(x)=f(x)，那么对任何常数显然也有[F(x)+C]'=f(x)，函数F(x)+C也是f(x)的原函数。这说明如果f(x)有一个原函数,那么f(x)就有无限多个原函数。
参考资料来源：百度百科——不定积分
热心网友
时间：2023-10-10 14:33
J(m)
=∫(0->π) x.(sinx)^m dx
let
u=π-x
= -dx
x=0, u=π
x=π, u=0
∫(0->π) x.(sinx)^m dx
=∫(π->0) (π-u) .(sinu)^m (-)
=∫(0->π) (π-x) .(sinx)^m dx
2∫(0->π) x.(sinx)^m dx =π∫(0->π) (sinx)^m dx
∫(0->π) x.(sinx)^m dx =(π/2)∫(0->π) (sinx)^m dx
consider
I(m)
=∫(0->π) (sinx)^m dx
=-∫(0->π) (sinx)^(m-1) dcosx
=-[ (sinx)^(m-1).cosx]|(0->π) +(m-1)∫(0->π)(cosx)^2.(sinx)^(m-2) dx
=(m-1)∫(0->π)(cosx)^2.(sinx)^(m-2) dx
=(m-1)∫(0->π)( 1-(sinx)^2).(sinx)^(m-2) dx
2∫(0->π) (sinx)^m dx =(m-1)∫(0->π) (sinx)^(m-2) dx
I(m) = [(m-1)/2] I(m-2)
I(1)=2
I(0) =π
case 1: m 是偶数
I(m)
= [(m-1)/2] I(m-2)
=[(1.3.5...m)/2^(m/2)] I(0)
=[(1.3.5...m)/2^(m/2)] π
J(m)
=∫(0->π) x.(sinx)^m dx
=(π/2)∫(0->π) (sinx)^m dx
=(π/2) I(m)
=(π/2) . { [(1.3.5...m)/2^(m/2)] π }
case 2 : m 是奇数
I(m)
= [(m-1)/2] I(m-2)
={ [(1.3.5...m)/2^[(m-1)/2] } I(1)
={ [(1.3.5...m)/2^[(m-1)/2] } )∫(0->π) (sinx) dx
={ [(1.3.5...m)/2^[(m-1)/2] } . 2
J(m)
=(π/2) I(m)
=π { [(1.3.5...m)/2^[(m-1)/2] }
扩展资料
常用积分公式：
1）∫0dx=c
2）∫x^udx=(x^(u+1))/(u+1)+c
3）∫1/xdx=ln|x|+c
4）∫a^xdx=(a^x)/lna+c
5）∫e^xdx=e^x+c
6）∫sinxdx=-cosx+c
7）∫cosxdx=sinx+c
8）∫1/(cosx)^2dx=tanx+c
9）∫1/(sinx)^2dx=-cotx+c
10）∫1/√（1-x^2) dx=arcsinx+c
热心网友
时间：2023-10-10 14:33
简单计算一下即可，答案如图所示
热心网友
时间：2023-10-10 14:34
=∫(0->π) x.(sinx)^m dx
let
u=π-x
= -dx
x=0, u=π
x=π, u=0
∫(0->π) x.(sinx)^m dx
=∫(π->0) (π-u) .(sinu)^m (-)
=∫(0->π) (π-x) .(sinx)^m dx
2∫(0->π) x.(sinx)^m dx =π∫(0->π) (sinx)^m dx
∫(0->π) x.(sinx)^m dx =(π/2)∫(0->π) (sinx)^m dx
consider
I(m)
=∫(0->π) (sinx)^m dx
=-∫(0->π) (sinx)^(m-1) dcosx
=-[ (sinx)^(m-1).cosx]|(0->π) +(m-1)∫(0->π)(cosx)^2.(sinx)^(m-2) dx
=(m-1)∫(0->π)(cosx)^2.(sinx)^(m-2) dx
=(m-1)∫(0->π)( 1-(sinx)^2).(sinx)^(m-2) dx
2∫(0->π) (sinx)^m dx =(m-1)∫(0->π) (sinx)^(m-2) dx
I(m) = [(m-1)/2] I(m-2)
I(1)=2
I(0) =π
case 1: m 是偶数
I(m)
= [(m-1)/2] I(m-2)
=[(1.3.5...m)/2^(m/2)] I(0)
=[(1.3.5...m)/2^(m/2)] π
J(m)
=∫(0->π) x.(sinx)^m dx
=(π/2)∫(0->π) (sinx)^m dx
=(π/2) I(m)
=(π/2) . { [(1.3.5...m)/2^(m/2)] π }
case 2 : m 是奇数
I(m)
= [(m-1)/2] I(m-2)
={ [(1.3.5...m)/2^[(m-1)/2] } I(1)
={ [(1.3.5...m)/2^[(m-1)/2] } )∫(0->π) (sinx) dx
={ [(1.3.5...m)/2^[(m-1)/2] } . 2
J(m)
=(π/2) I(m)
=π { [(1.3.5...m)/2^[(m-1)/2] }
