Integration of a function which has limits in terms of parameters
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aakash dewangan
le 12 Mai 2021
Commenté : Walter Roberson
le 16 Mai 2021
Hi,
I have a function f3(t), defined as
clc; clear all;
syms t L rho m L n T k G v
N(t) = (T*L/2)*(n*pi/L)^2+k*sin(n*pi*v*t/L);
D(t) = (rho*L/2+m*(sin(n*pi*v*t/L))^2);
alpha(t) = N/D;
i = 6; j = 1
f3(t) = alpha(t)*cos(2*pi*i*t/L)*sin(2*pi*j*t/L);
I3 = int(f3,t,0,L)
when i am integrating the function f3(t) from limits 0 to L, the code is not giving me the output in the form of some expression. The output i am getting is in the form of int(f3,t,0,L), which I dont want. I want output in the form of expression (which will contain t L rho m L n T k G v).
Please help me with this!! Any help will be appreciated..
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Walter Roberson
le 16 Mai 2021
I have no suggestions.
There just might be a change of variables available to make sin(n*pi*v*t/L) linear.
Is n*v known to be integer? If so then that would make a difference in the integration, as sin(n*pi*v*t/L) at t=L would be sin(n*pi*v) and if n*v were integer that would be 0 .
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Jan
le 14 Mai 2021
syms t L rho m L n T k G v
N(t) = (T*L/2)*(n*pi/L)^2+k*sin(n*pi*v*t/L);
D(t) = (rho*L/2+m*(sin(n*pi*v*t/L))^2);
alpha(t) = N/D;
i = 6; j = 1;
f3(t) = alpha(t)*cos(2*pi*i*t/L)*sin(2*pi*j*t/L);
I3 = int(f3, t, [0,L])
As Walter has said already: There is no closed form solution of this integral. You can simplify it only, if you provide specific values for the parameters.
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