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

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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.

##### 1 Comment

### Accepted Answer

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()

##### 1 Comment

Walter Roberson
on 5 May 2021

Edited: 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)

z0 = limit(z, t, 0, 'right');

char(z0)

res = simplify(z - z0);

char(res)

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.

### More Answers (2)

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)

z = int(mag_dr, t)

z - limit(z, t, 0, 'right')

The integral is discontinuous at 0, which is why it cannot be resolved by MATLAB.

##### 4 Comments

Dyuman Joshi
on 6 May 2021

Walter Roberson
on 6 May 2021

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.

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