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Tika Ram Pokhrel on 2 May 2021
Commented: Walter Roberson on 6 May 2021
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.
##### 2 CommentsShowHide 1 older comment
Walter Roberson on 5 May 2021
Integration by parts, using a change of variables u=4*t

Dyuman Joshi on 5 May 2021
Edited: Dyuman Joshi on 6 May 2021
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.
Walter Roberson on 5 May 2021
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.

Walter Roberson on 5 May 2021
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.
Walter Roberson on 6 May 2021
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.

Sindhu Karri on 5 May 2021
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.
Walter Roberson on 5 May 2021
It does exist.

R2020b

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