How do i further simplify this solution in matlab. I have used the simplify(.....) but it still gives me the same solution with the sin functions.

Later you assign a value to L, and hence it works on integration, however you still have its symbolic form in z1, and hence it returns an answer dependent on L

In short, you arent getting a simplified answer because your answer is an expression with a variable.

Try this:

syms z
syms z1
L = 33.65;
z1= 0.5*L;
pi = 3.1412;
valueBxx1=int(sin(pi*z1/L)^2 ,z, 0, L);
vpa(simplify(valueBxx1),3)

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George Adomako Kumi 2022-11-1

thanks

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Answer 2

John D'Errico 2022-11-1

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编辑：John D'Errico 2022-11-1

在 MATLAB Online 中打开

ARGH!!!!!! Why would you define the value of pi in MATLAB? Worse, then not even defining it to be the correct value?

pi already exists as a variable in MATLAB.

format long
pi
ans = 
   3.141592653589793

3.1412 is not even correctly rounded!

And worse yet, you are doing symbolic computations!!!!! Why spend the computing power to compute the wrong thing exactly?????

(Excuse me, as I go pound my head against a local brick wall. Ok. Now my head feels much better.)

syms z L z1 za

z1 = 0.5*L;

L = 33.65;

valueBxx1 = int(sin(pi*z1/L)^2 ,z, 0, L)

valueBxx1 =

What would you want to see there? L is unknown, and it is inside the trig function. The sin of that expression is a FUNCTION OF L. Trig functions are NONLINEAR functions. As such, they have the fundamental property that they CANNOT be simplified. There is little more that could be done with it, since the value of L is unknown.

Could you force MATLAB to rewrite that expression using say a double angle formula for sin(x)^2? So I suppose we could use an identity like:

sin(u)^2 = (1 - (cos(2*u))/2

It will still have a cosine function of L. What more would you want to see?

simplify(valueBxx1,'steps',10,'all',true)

ans =

So the second of those choices does not have the square of a sine function in it. Is it materially different?