Given \[ f(x)=\int_{0}^{x} \frac{t^{2}-49}{1+\cos ^{2}(t)} d t \] At what value of $x$ does the local max of $f(x)$ occur?

Step 1 ：Given the function \(f(x)=\int_{0}^{x} \frac{t^{2}-49}{1+\cos ^{2}(t)} d t\), we need to find the value of \(x\) at which the local maximum of \(f(x)\) occurs.

Step 2 ：The local maximum of a function occurs where the derivative of the function is zero and the second derivative is negative.

Step 3 ：First, we find the derivative of \(f(x)\), which is \((x - 7)*(x + 7)/(\cos(x)^{2} + 1)\).

Step 4 ：We set the derivative equal to zero and solve for \(x\), giving us the critical points \([-7, 7]\).

Step 5 ：Next, we find the second derivative of \(f(x)\), which is \(2*(x - 7)*(x + 7)*\sin(x)*\cos(x)/(\cos(x)^{2} + 1)^{2} + (x - 7)/(\cos(x)^{2} + 1) + (x + 7)/(\cos(x)^{2} + 1)\).

Step 6 ：We evaluate the second derivative at the critical points, giving us the values \([-8.92647297244674, 8.92647297244674]\).

Step 7 ：The second derivative at \(x = -7\) is negative and at \(x = 7\) is positive. This means that the function has a local maximum at \(x = -7\) and a local minimum at \(x = 7\).

Step 8 ：Therefore, the local maximum of \(f(x)\) occurs at \(x = -7\).

Step 9 ：Final Answer: The local maximum of \(f(x)\) occurs at \(\boxed{-7}\).