The given integral is:

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} -\cos x \, dx$

Step-by-step solution:

1. Factor out the negative sign:

   $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} -\cos x \, dx = - \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x \, dx$

2. Find the antiderivative of $\cos x$:

   The antiderivative of $\cos x$ is $\sin x$.

   $\int \cos x \, dx = \sin x$

3. Evaluate the definite integral:

   Substitute the limits $-\frac{\pi}{2}$ and $\frac{\pi}{2}$:

   $- \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x \, dx = - \left[ \sin x \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$

   Compute $\sin x$ at the limits:

   $\sin \left(\frac{\pi}{2}\right) = 1, \quad \sin \left(-\frac{\pi}{2}\right) = -1$

   Therefore:

   $\left[ \sin x \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = \sin \left(\frac{\pi}{2}\right) - \sin \left(-\frac{\pi}{2}\right) = 1 - (-1) = 2$

4. Multiply by the negative sign:

   $- \left[ \sin x \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = -2$

Final Answer:

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} -\cos x \, dx = -2$

Would you like a more detailed explanation or clarification?

Here are some related questions you might find interesting:

1. What is the significance of negative values in definite integrals?
2. How does symmetry help in evaluating definite integrals over symmetric intervals?
3. How would the answer change if the integrand were $|\cos x|$ instead of $-\cos x$?
4. Can you generalize this result to other trigonometric functions like $\sin x$?
5. How can definite integrals be applied to solve real-world problems?

Tip: Always check for symmetries in definite integrals, as they can simplify the computation significantly!