cos(x)/1-a*cos(x)对x积分
时间: 2024-06-01 16:12:18 浏览: 35
首先,我们可以通过分离分子分母的方法将被积函数变形:
cos(x)/(1-a*cos(x)) = 1/(a-1) * [ (a-1)*cos(x)/(1-a*cos(x)) + 1 ]
然后,我们可以对变形后的被积函数分别进行积分:
∫cos(x)/(1-a*cos(x)) dx = 1/(a-1) * [ ∫(a-1)*cos(x)/(1-a*cos(x)) dx + ∫1 dx ]
对第一项进行有理分式分解,得到:
(a-1)*cos(x)/(1-a*cos(x)) = 1 + (a-1)/(a-1*cos(x)-1)
再进行变量代换,令 u = a-1*cos(x)-1,则 du = (a-1)*sin(x) dx,从而有:
∫(a-1)*cos(x)/(1-a*cos(x)) dx = -∫(a-1)/(u) du = -(a-1)*ln|a-1*cos(x)-1| + C1
对第二项进行积分,得到:
∫1 dx = x + C2
因此,原函数为:
∫cos(x)/(1-a*cos(x)) dx = 1/(a-1) * [ -(a-1)*ln|a-1*cos(x)-1| + x ] + C
其中 C1,C2,C 为常数。
相关问题
syms x; syms y; syms a; syms b; syms c; x0=-pi/4;x1=pi/4; y0=-tan(c)*tan(23.43*pi/180); y1=tan(c)*tan(23.43*pi/180); fun1 = sin(c)*(sin(x)-x*cos(x))/(cos(x)^2+tan(c)^2)^0.5 fun2 = (0.506-0.476*cos(y))*sin(x-1/3*pi)+0.6609*cos(y)+0.409 fun3 = (cos(y)-cos(x))/(sin(x)-2*pi/360*x*cos(x)) fun41 = sin(c)*(cos(y)-cos(x))/(cos(x)^2+tan(c)^2)^0.5 fun42 = 2*asin(fun41)-2*a-b fun4 = 0.5*cos(fun42)+0.5*cos(a+b) f = fun1*fun2*fun3*fun4 gongshi=int(f,y,y0,y1) jieguo=int(gongshi,x,x0,x1)
根据您的描述,这是一段 Matlab 代码,其中定义了多个符号变量和函数,并进行了积分计算。根据您的代码,我们可以按照以下步骤进行计算:
1. 定义符号变量和函数:
```
syms x y a b c;
x0=-pi/4;
x1=pi/4;
y0=-tan(c)*tan(23.43*pi/180);
y1=tan(c)*tan(23.43*pi/180);
fun1 = sin(c)*(sin(x)-x*cos(x))/(cos(x)^2+tan(c)^2)^0.5;
fun2 = (0.506-0.476*cos(y))*sin(x-1/3*pi)+0.6609*cos(y)+0.409;
fun3 = (cos(y)-cos(x))/(sin(x)-2*pi/360*x*cos(x));
fun41 = sin(c)*(cos(y)-cos(x))/(cos(x)^2+tan(c)^2)^0.5;
fun42 = 2*asin(fun41)-2*a-b;
fun4 = 0.5*cos(fun42)+0.5*cos(a+b);
```
2. 计算被积函数:
```
f = fun1*fun2*fun3*fun4;
```
3. 对 y 进行一次积分:
```
gongshi = int(f, y, y0, y1);
```
4. 对 x 进行一次积分:
```
jieguo = int(gongshi, x, x0, x1);
```
5. 输出积分结果:
```
disp(jieguo);
```
这样就可以得到积分结果了。需要注意的是,如果符号变量中的参数 $a,b,c$ 没有被赋值,那么积分结果中仍然会包含这些参数,需要根据具体情况进行替换。
计算ln(1+1/x)/(sinx*(cos2x+1))的不定积分
首先,化简被积函数:
ln(1 1/x) = ln(x/(x-1)) = ln x - ln(x-1)
cos2x 1 = (1 + cos2x)/2 = (1 + 2cos2x - 1)/2 = cos2x + 1/2
因此,被积函数可以写成:
(ln x - ln(x-1))/(sinx*(cos2x + 1/2))
接下来,分部积分法:
设f(x) = ln x - ln(x-1),g(x) = 1/(sinx*(cos2x + 1/2))
f'(x) = 1/x - 1/(x-1),g'(x) = (-sinx*cos2x - sinx)/(sinx^2*(cos2x + 1/2)^2)
积分式可以写成:
∫f(x)g'(x)dx = f(x)g(x) - ∫g(x)f'(x)dx
计算f(x)g(x):
f(x)g(x) = (ln x - ln(x-1))/(sinx*(cos2x + 1/2))
计算∫g(x)f'(x)dx:
∫g(x)f'(x)dx = ∫(-sinx*cos2x - sinx)/(sinx^2*(cos2x + 1/2)^2)*(1/x - 1/(x-1))dx
化简:
∫g(x)f'(x)dx = ∫(-cos2x - 1/2)/(sinx*(cos2x + 1/2)^2)dx + ∫(cos2x + 1/2)/(sinx^2*(cos2x + 1/2)^2)dx
对于第一个积分,令u = cos2x + 1/2,du = -2sin2xdx,得到:
∫(-cos2x - 1/2)/(sinx*(cos2x + 1/2)^2)dx = -∫du/u^2 = 1/u + C = 2/(2cos2x + 1) + C
对于第二个积分,可以使用三角恒等式sin2x = 2sinx*cosx:
∫(cos2x + 1/2)/(sinx^2*(cos2x + 1/2)^2)dx = ∫(cos2x + 1/2)/(2sinx*cos2x)^2*(2cosx)dx
令u = cos2x,du = -sin2xdx,得到:
∫(cos2x + 1/2)/(sinx^2*(cos2x + 1/2)^2)dx = -1/sin2x + ∫du/(2u^2) = -1/(2sinxcos2x) + 1/(4cos2x) + C
将f(x)g(x)和∫g(x)f'(x)dx代入分部积分公式,得到:
∫(ln x - ln(x-1))/(sinx*(cos2x + 1/2))dx = (ln x - ln(x-1))*[2/(2cos2x + 1)] - [2/(2sinxcos2x) - 1/(4cos2x)] + C
化简可得:
∫(ln x - ln(x-1))/(sinx*(cos2x + 1/2))dx = ln(x*(2cos2x + 1)/(2sinxcos2x*(x-1))) + 1/(4cos2x) + C
因此,原式的不定积分为:
ln(x*(2cos2x + 1)/(2sinxcos2x*(x-1))) + 1/(4cos2x) + C
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