I could not integrate using MatLab, Can you please help me?

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Tika Ram Pokhrel
Tika Ram Pokhrel 2021-5-2
评论： Walter Roberson 2021-5-6
In solving a problem I need to integrate the following function with respect to 't' from the limit 0 to t.
3*2^(1/2)*(1 - cos(4*t))^(1/2)*(a^2 + c^2)^(1/2)
I used the following commands but got the same result as given herewith.
>> syms a c t real
mag_dr = 3*2^(1/2)*(1 - cos(4*t))^(1/2)*(a^2 + c^2)^(1/2)
>> int(mag_dr,t,0,t)
ans =
int(3*2^(1/2)*(1 - cos(4*t))^(1/2)*(a^2 + c^2)^(1/2), t, 0, t)
Let me know the best way(s) to tackle this type of problem.

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采纳的回答

Dyuman Joshi
Dyuman Joshi 2021-5-5
编辑：Dyuman Joshi 2021-5-6
syms t a c
fun = 3*2^(1/2)*(1 - cos(4*t))^(1/2)*(a^2 + c^2)^(1/2);
z = int(fun,t); %gives indefinite integral
%result of integration, z = -(3*sin(4*t)*(a^2 + c^2)^(1/2))/(2*(sin(2*t)^2)^(1/2));
t=0;
res = z - subs(z);
%obtain final result by evaluating the integral, z(t)-z(0), by assigning t & using subs()
However, you will not get the result. See @Walter Roberson's comment below for more details.
  1 个评论
Walter Roberson
Walter Roberson 2021-5-5
编辑：Walter Roberson 2021-5-5
Ran in:
Not quite.
syms t a c
fun = 3*2^(1/2)*(1 - cos(4*t))^(1/2)*(a^2 + c^2)^(1/2);
z = int(fun,t); %gives indefinite integral
char(z)
ans = '-(3*sin(4*t)*(a^2 + c^2)^(1/2))/(2*(sin(2*t)^2)^(1/2))'
z0 = limit(z, t, 0, 'right');
char(z0)
ans = '-3*(a^2 + c^2)^(1/2)'
res = simplify(z - z0);
char(res)
ans = '3*(a^2 + c^2)^(1/2) - (3*sin(4*t)*(a^2 + c^2)^(1/2))/(2*(sin(2*t)^2)^(1/2))'
fplot(subs(z, [a,c], [1 2]), [-5 5])
fplot((subs(fun,[a,c], [1 2])), [-5 5])
That is, the problem is that the integral is discontinuous at t = 0 and that is why int() cannot resolve it.

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更多回答（2 个）

Walter Roberson
Walter Roberson 2021-5-5
Ran in:
syms a c t real
mag_dr = 3*2^(1/2)*(1 - cos(4*t))^(1/2)*(a^2 + c^2)^(1/2)
mag_dr = 
z = int(mag_dr, t)
z = 
z - limit(z, t, 0, 'right')
ans = 
The integral is discontinuous at 0, which is why it cannot be resolved by MATLAB.
  4 个评论
显示 2更早的评论
Dyuman Joshi
Dyuman Joshi 2021-5-6
The wrong substitution was a mistake on my part, mostly cause I did it in a hurry. I have edited my nswer accordingly as well. Other than that, is subs() a good approach or would you recommend otherwise?
Walter Roberson
Walter Roberson 2021-5-6
limit() is more robust than subs() for cases like this. But limit() is sometimes quite expensive to calculate, or is beyond MATLAB's ability to calculate, even in some finite cases.

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Sindhu Karri
Sindhu Karri 2021-5-5
Hii
The "int" function cannot solve all integrals since symbolic integration is such a complicated task. It is also possible that no analytic or elementary closed-form solution exists.
For definite integrals, a numeric approximation can be performed by using the "integral" function.

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