主题:帮帮忙吧,大侠!
用SIMPSON法对含有K,P的式子积分,K,P是未知常量,但积出来是常数怎么回事?
%%%%SIMPSON的M文件
function rs=simpson(s,x,a,b,n)
h=(b-a)/n;
r=feval(s,a)+feval(s,b);
for j=1:2:n-1
x=a+j*h
r=r+4*feval(s,x);
end
for j=2:2:n-2
x=a+j*h;
r=r+2*feval(s,x);
end
rs=r*h/3
要积分的式子:(p/(1-k*cos(4*x))+15*p/(1-k*cos(4*x))*(16*p^2/(1-k*cos(4*x))^4*k^2*sin(4*x)^2-p/(1-k*cos(4*x))*(32*p/(1-k*cos(4*x))^3*k^2*sin(4*x)^2-16*p/(1-k*cos(4*x))^2*k*cos(4*x)))/((16*p^2/(1-k*cos(4*x))^4*k^2*sin(4*x)^2+p^2/(1-k*cos(4*x))^2)^3)^(1/2))/(p/(1-k*cos(4*x))+15*p/(1-k*cos(4*x))/(16*p^2/(1-k*cos(4*x))^4*k^2*sin(4*x)^2+p^2/(1-k*cos(4*x))^2)^(1/2))
%%%%SIMPSON的M文件
function rs=simpson(s,x,a,b,n)
h=(b-a)/n;
r=feval(s,a)+feval(s,b);
for j=1:2:n-1
x=a+j*h
r=r+4*feval(s,x);
end
for j=2:2:n-2
x=a+j*h;
r=r+2*feval(s,x);
end
rs=r*h/3
要积分的式子:(p/(1-k*cos(4*x))+15*p/(1-k*cos(4*x))*(16*p^2/(1-k*cos(4*x))^4*k^2*sin(4*x)^2-p/(1-k*cos(4*x))*(32*p/(1-k*cos(4*x))^3*k^2*sin(4*x)^2-16*p/(1-k*cos(4*x))^2*k*cos(4*x)))/((16*p^2/(1-k*cos(4*x))^4*k^2*sin(4*x)^2+p^2/(1-k*cos(4*x))^2)^3)^(1/2))/(p/(1-k*cos(4*x))+15*p/(1-k*cos(4*x))/(16*p^2/(1-k*cos(4*x))^4*k^2*sin(4*x)^2+p^2/(1-k*cos(4*x))^2)^(1/2))