Suppose that $f(x)=\int_{0}^{x} \frac{t^{2}-25}{1+\cos ^{2}(t)} d t$. At what value of $x$ does the local maximum of $f(x)$ occur? At what value of $x$ does the local minimum of $f(x)$ occur? \[ x= \]

Step 1 ：Suppose that \(f(x)=\int_{0}^{x} \frac{t^{2}-25}{1+\cos ^{2}(t)} d t\). We need to find the local maximum and minimum of this function.

Step 2 ：The local maximum and minimum of a function occur where its derivative is zero. So, we need to find the derivative of the function \(f(x)\) and set it equal to zero.

Step 3 ：The derivative of an integral from a constant to a variable is just the integrand evaluated at that variable. So, the derivative of \(f(x)\) is \(\frac{x^{2}-25}{1+\cos ^{2}(x)}\).

Step 4 ：We need to solve the equation \(\frac{x^{2}-25}{1+\cos ^{2}(x)} = 0\) for \(x\). The solutions to this equation are \(x = -5\) and \(x = 5\).

Step 5 ：However, these are just the points where the derivative of the function is zero. We need to determine which of these points is a local maximum and which is a local minimum.

Step 6 ：We can do this by taking the second derivative of the function and evaluating it at these points. If the second derivative is positive at a point, then that point is a local minimum. If the second derivative is negative at a point, then that point is a local maximum.

Step 7 ：The second derivative of the function is \(2x/(\cos(x)^{2} + 1) + 2(x^{2} - 25)\sin(x)\cos(x)/(\cos(x)^{2} + 1)^{2}\).

Step 8 ：Evaluating the second derivative at \(x = -5\) gives a negative value, and evaluating it at \(x = 5\) gives a positive value.

Step 9 ：Therefore, the function \(f(x)\) has a local maximum at \(x = -5\) and a local minimum at \(x = 5\).

Step 10 ：Final Answer: The local maximum of \(f(x)\) occurs at \(x = \boxed{-5}\) and the local minimum of \(f(x)\) occurs at \(x = \boxed{5}\).