I want to evaluate the integral of this function :
(cos(t^2)*sin(t)^2)^(1/2)
over the period
-15*pi:50*pi
I am using these :
f=@(t) (cos(t.^2).*sin(t).^2).^(1/2);
quad(f,-15*pi,50*pi)
But i am getting warned:
Warning: Maximum function count exceeded; singularity likely.
> In quad at 110
In PPEL at 6
ans =
35.8252 +54.5673i
Most likely there is something stupid that i don't know about the theoretical part of the math, like some singularity as MATLAB said.
Moreover, when i want to evaluate this function sybolically with this :
int((cos(t.^2).*sin(t).^2).^(1/2))
It says : Warning: Explicit integral could not be found.
ans =
int((cos(t^2)*sin(t)^2)^(1/2), t)
What's the solution ?
Plot the function without trying to do the integral.
What I notice is that you are going to be taking the square root of a negative number. I suspect that that isn't what you want to be doing...