Help!: int(expr,var) function returns the int(expr,var) function

1 visualizzazione (ultimi 30 giorni)
When i try to make a fucntion described by two indefinite integrals, one of the indefinite integrals returns without being solved.
The variable x an angle between 0 and pi/2.
F_1 is a force.
R_1 and R_2 are lenghts.
%% Variables
syms x;
F_1 = 200; %N
R_1 = 30; % mm
L = 40; % mm
H = 20; % mm
R_2 = 20*(4*cos(x)^2 + sin(x)^2)^(1/2);
%% Equations
m_1(x)= F_1*sin(x)*R_1;
md_1(x)=sin(x)*R_1;
m_2(x)=F_1*R_2*cos(x);
md_2(x)=R_2*cos(x);
ex_1=R_1*m_1(x)*md_1(x);
ex_2=simplify(R_2*m_2(x)*md_2(x))
%% Integration
f(x)=int(ex_1,x) + int(ex_2,x)
f(x) =
2700000*x - 1350000*sin(2*x) - int(1600000*(4 - 3*sin(x)^2)^(3/2)*(sin(x)^2 - 1), x) - It should solve this one, but indstead it just spit out int(expresion,var) just as it was inserted.

Risposta accettata

Cris LaPierre
Cris LaPierre il 17 Apr 2021
That likely means it couldn't solve that one.
Consider using one of the techniques mentioned in these two examples.
  1 Commento
Daniel Møgelbjerg
Daniel Møgelbjerg il 20 Apr 2021
We did it a little different, after realizing that our R_2 in fact was not a variable depending on x, but just another constant value. We are sorry for the eventually wasted time, but thanks a lot for the help.

Accedi per commentare.

Più risposte (0)

Prodotti


Release

R2020a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by