0

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 ?

Is your function really supposed to be the square root of (the multiplication of the (cos of the square of t) and the (square of the sin of t)) - Mark Elliot 2012-04-04 02:31

2

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...