# Integral "int" function not evaluating

2 views (last 30 days)
Ali Almakhmari on 3 Mar 2022
Commented: Walter Roberson on 6 Mar 2022
Why is the "int" function not evaluating my integral? The "r" variable still has the "int" in it.
clear
clc
theta_s = 0:0.5:pi/2;
syms theta_v phi
for i = 1:length(theta_s)
integ2(i) = ((((1/(2*pi))*((pi-phi)*cos(phi) + sin(phi))*tan(theta_s(i))*tan(theta_v)-(1/pi)*(tan(theta_s(i))+tan(theta_v)+sqrt(tan(theta_v)^2 + tan(theta_s(i))^2 - 2*tan(theta_s(i))*tan(theta_v)*cos(phi)))))*cos(theta_v)*sin(theta_v));
r(i)= int((integ2(i)), theta_v, [0 pi/2]);
end
Walter Roberson on 3 Mar 2022
What reason do you have to lead you to expect that there is a closed form integral?

David Hill on 3 Mar 2022
theta_s = 0:0.5:pi/2;
syms theta_v
phi=pi/6;%choose a phi or loop for various values of phi
for i = 1:length(theta_s)
integ2(i) = ((((1/(2*pi))*((pi-phi)*cos(phi) + sin(phi))*tan(theta_s(i))*tan(theta_v)-(1/pi)*(tan(theta_s(i))+tan(theta_v)+sqrt(tan(theta_v)^2 + tan(theta_s(i))^2 - 2*tan(theta_s(i))*tan(theta_v)*cos(phi)))))*cos(theta_v)*sin(theta_v));
r(i)= vpaintegral((integ2(i)), theta_v, [0 pi/2]);
end
Walter Roberson on 6 Mar 2022
By the way, integrating first with respect to phi is faster.
At one point I saw a closed form solution for integration with respect to phi, but I was not able to reproduce that later.

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