Integral "int" function not evaluating

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]

What reason do you have to lead you to expect that there is a closed form integral?

Answer 1

[David Hill]

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

[Ali Almakhmari]

But I need an answer in terms of phi

[Torsten]

If "int" does not return an analytical expression for the integral in terms of phi, you'll have to be satisfied with an evaluation for different numerical values of phi.

[AndresVar]

@Ali Almakhmari the square root part is too complicated to find a simplified closed form. Could there be a typo or maybe use some identity to simplify?

[Ali Almakhmari]

Actually I forgot to say this, but after integrating this function in terms of theta_v, I will do another integration in terms of phi from 0 to pi, so am not looking for a symbolic answer at the end if that helps. I just don't how to do it because matlab is not evaulating it even with two int functions (for theta_v and phi)

[David Hill]

Works fine.

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)= vpaintegral(vpaintegral((integ2(i)), theta_v, [0 pi/2]),phi,[0 pi]);
end

[Walter Roberson]

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.