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

조회 수: 24 (최근 30일)
Tika Ram Pokhrel
Tika Ram Pokhrel 2021년 5월 2일
댓글: Walter Roberson 2021년 5월 6일
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.

채택된 답변

Dyuman Joshi
Dyuman Joshi 2021년 5월 5일
편집: Dyuman Joshi 2021년 5월 6일
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.
  댓글 수: 1
Walter Roberson
Walter Roberson 2021년 5월 5일
편집: Walter Roberson 2021년 5월 5일
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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추가 답변 (2개)

Walter Roberson
Walter Roberson 2021년 5월 5일
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.
  댓글 수: 4
Dyuman Joshi
Dyuman Joshi 2021년 5월 6일
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 2021년 5월 6일
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 2021년 5월 5일
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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