大一高数定积分与不定积分求解
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/08/29 05:42:35
大一高数定积分与不定积分求解
![](http://img.wesiedu.com/upload/1/c7/1c78d88d2c8322963fb864175cfa1b6f.jpg)
![](http://img.wesiedu.com/upload/1/c7/1c78d88d2c8322963fb864175cfa1b6f.jpg)
本题是三角函数定积分的经典问题,推导过程如下
作变量置换 y = x - π/2,则x = y + π/2,原积分式化为:
[0,π]∫x*(sinx)^n *dx
= [-π/2,π/2]∫(y+π/2)*(sin(y+π/2))^n *dy
= [-π/2,π/2]∫y*(cosy)^n *dy + [-π/2,π/2]∫π/2*(cosy)^n *dy
显然和式第一项被积函数为奇函数,因此第一项积分结果为0;
和式第二项被积函数为偶函数,因此对称区间等于2倍的半区间积分,原式化为:
原式= 2* [0,π/2]∫π/2*(cosy)^n *dy
= π* [0,π/2]∫(cosy)^n *dy
再作变换 u =π/2 –y ==> du =-dy,有:
π* [0,π/2]∫(cosy)^n *dy
= π* [π/2,0]∫(cos(π/2 –u)^n *(-du)
=π* [0,π/2]∫(sinu)^n *du
因此:
[0,π]∫x*(sinx)^n *dx =π* [0,π/2]∫(sinx)^n *d =π* [0,π/2]∫(cosx)^n *dx
设 F(n) = [0,π/2]∫(sinx)^n *dx,则有:
F(n) = [0,π/2]∫(sinx)^n *dx
= [0,π/2]∫(sinx)^(n-1) *d(-cosx)
= - [0,π/2]|(sinx)^n*(-cosx) + [0,π/2]∫(n-1)(sinx)^(n-2) *cos²x *dx
= 0 + [0,π/2]∫(n-1)(sinx)^(n-2) *(1-sin²x) *dx
= (n-1) * [0,π/2]∫(sinx)^(n-2) - (n-1)* [0,π/2]∫(sinx)^n *dx
= (n-1)*F(n-2) – (n-1)F(n)
解得:F(n) = (n-1)/n *F(n-2)
F(1) = [0,π/2]∫sinx *dx = 1;F(0) = [0,π/2]∫dx =π/2
因此按照递推公式得到:
当n为偶数时:
F(n) = (n-1)/n * (n-3)/(n-2)*…..*1/2* F(0) = (n-1)/n * (n-3)/(n-2)*…..*1/2*π/2
当 n为奇数时:
F(n) = (n-1)/n * (n-3)/(n-2)*…..*2/3* F(1) = (n-1)/n * (n-3)/(n-2)*…..*2/3* 1
因此:
[0,π]∫x*(sinx)^n *dx =
(1) π* (n-1)/n * (n-3)/(n-2)*…..*1/2*π/2 (n为偶数)
(2) π* (n-1)/n * (n-3)/(n-2)*…..*2/3*1 (n为奇数)
作变量置换 y = x - π/2,则x = y + π/2,原积分式化为:
[0,π]∫x*(sinx)^n *dx
= [-π/2,π/2]∫(y+π/2)*(sin(y+π/2))^n *dy
= [-π/2,π/2]∫y*(cosy)^n *dy + [-π/2,π/2]∫π/2*(cosy)^n *dy
显然和式第一项被积函数为奇函数,因此第一项积分结果为0;
和式第二项被积函数为偶函数,因此对称区间等于2倍的半区间积分,原式化为:
原式= 2* [0,π/2]∫π/2*(cosy)^n *dy
= π* [0,π/2]∫(cosy)^n *dy
再作变换 u =π/2 –y ==> du =-dy,有:
π* [0,π/2]∫(cosy)^n *dy
= π* [π/2,0]∫(cos(π/2 –u)^n *(-du)
=π* [0,π/2]∫(sinu)^n *du
因此:
[0,π]∫x*(sinx)^n *dx =π* [0,π/2]∫(sinx)^n *d =π* [0,π/2]∫(cosx)^n *dx
设 F(n) = [0,π/2]∫(sinx)^n *dx,则有:
F(n) = [0,π/2]∫(sinx)^n *dx
= [0,π/2]∫(sinx)^(n-1) *d(-cosx)
= - [0,π/2]|(sinx)^n*(-cosx) + [0,π/2]∫(n-1)(sinx)^(n-2) *cos²x *dx
= 0 + [0,π/2]∫(n-1)(sinx)^(n-2) *(1-sin²x) *dx
= (n-1) * [0,π/2]∫(sinx)^(n-2) - (n-1)* [0,π/2]∫(sinx)^n *dx
= (n-1)*F(n-2) – (n-1)F(n)
解得:F(n) = (n-1)/n *F(n-2)
F(1) = [0,π/2]∫sinx *dx = 1;F(0) = [0,π/2]∫dx =π/2
因此按照递推公式得到:
当n为偶数时:
F(n) = (n-1)/n * (n-3)/(n-2)*…..*1/2* F(0) = (n-1)/n * (n-3)/(n-2)*…..*1/2*π/2
当 n为奇数时:
F(n) = (n-1)/n * (n-3)/(n-2)*…..*2/3* F(1) = (n-1)/n * (n-3)/(n-2)*…..*2/3* 1
因此:
[0,π]∫x*(sinx)^n *dx =
(1) π* (n-1)/n * (n-3)/(n-2)*…..*1/2*π/2 (n为偶数)
(2) π* (n-1)/n * (n-3)/(n-2)*…..*2/3*1 (n为奇数)