Why doesnt matlab simplify the calculations.How could i improve my code?

%In determining the load distribution in axial thrust bearings under an eccentric load,the following integral must be evaluated:

e=0.6;
m=0||1;
syms x

c=1.1;
a=(cos(1-2.*e));
na=-1.*(cos(1-2.*e));
eqn=@(x) (([1+(-1+acos(x))/(2.*e)].^(c)).*(cos(m.*x)));
Im=(1/(2*pi))*(int(eqn,x,na,a));
pretty(Im)

Answer 1

[Star Strider]

In your problem, there may not be an analytical expression for the integral. If you want the numerical value, use the vpa function:

e=0.6;
m=0.1;
syms x

c=1.1;
a=(cos(1-2.*e));
na=-1.*(cos(1-2.*e));
eqn=@(x) (([1+(-1+acos(x))/(2.*e)].^(c)).*(cos(m.*x)));
Im=(1/(2*pi))*(int(eqn,x,na,a));
Im = simplify(Im, 'Steps',10);                              % Symbolic Simplification
pretty(Im)
Im_val = vpa(Im)                                            % Numeric Value

Im_val =
0.48167541183902032831734712665393

I don’t know what you want for ‘m’. What you posted for it doesn’t make sense. Change it in my code to whatever it needs to be.

[Shane Mcfarlane]

In the problem it says m= 0 or 1. also the answer is I1(0.6) = 0.2416. Im not sure how they arrive at that .

[Walter Roberson]

To reach that integral with those equations, m would have to be about 1.9242

[Walter Roberson]

Alternately, with m = 1, the value 0.2416 would be reached with e approximately 0.96

Answer 2

[Walter Roberson]

int() is not documented as accepting a function handle, only a symbolic expression or symbolic function.