Hi, can any one please help me in solving the integral involving in this problem, The pdf is also attached. Thank you in advance

g = 1.55;

t = 2*10^-9;

L = 50*10^-6;

R = 50*10^-9;

u = (R+t)^2 - R^2;

kp = 3000;

kf = 0.663;

phi = 0.1;

phi1 = pi;

e = sqrt( 1 - (( R^2 + u ) / ( (L/2)^2 + u)));

G = ( 2 * e * ( 1 - e^2 )^1/6 ) / ( e * sqrt((1 - e^2)) + asin(e));

n = 3 * G^-g;

C = (( R + t )^2 - R^2 )/ R^2;

klr = ( kp * R * (1 + t/R - kf/ kp) * log (1 + t/R ))/ ( t * kf * log ((1 + t/R ) * kp / kf));

B = ( R/ (R + t))* ((( kp - klr)/(klr + kp)) - 1);

A = - (2 * klr / (kp + klr));

kpez = (kp + C * klr)/ (1 + C);

kpex = (A * phi * kp + B * C * phi *klr) / (A * phi + B * C * phi);

syms x;

g = inline('sqrt( (kpez)^2 * sin(x)^2+ (kpex)^2 * cos(x)^2)', 'x', 'kpez' , 'kpex')

kpe = (1/ pi) * int(g(x, kpez, kpex),x,0,pi);

B1 = (kpe + kf * (n - 1) + (n + 1)* (kpe - kf) * (1 + C) * phi1)/ (kpe + kf * (n - 1)- (kpe-kf) * (1 + C) * phi1 )

[Karan Gill]

What problem did you run into when you ran the code?

[zakir hussain]

Dear sir I am getting problem to calculate 'kpe' in second last line (integral).

Answer 1

[Torsten]

Remove

syms x

and use

g=@(x)sqrt(kpez^2*sin(x).^2+kpex^2*cos(x).^2);
kpe=1/pi*integral(g,0,pi);

Best wishes

Torsten.

[zakir hussain]

Thank you so much Torsten.