Evaluate the following integral using the Fundamental Theorem of Calculus, \( \int_{-5 \pi / 2}^{5 \pi / 2}(\cos x-2) d x \)

To evaluate the integral \( \int_{-5 \pi / 2}^{5 \pi / 2}(\cos x-2) d x \) using the Fundamental Theorem of Calculus, first identify an antiderivative of the integrand. The antiderivative of \( \cos x \) is \( \sin x \), and the antiderivative of \( -2 \) is \( -2x \). Therefore, an antiderivative \( F(x) \) of \( \cos x - 2 \) is: \[ F(x) = \sin x - 2x \] Next, we will evaluate \( F(x) \) at the limits of integration from \( -5 \pi / 2 \) to \( 5 \pi / 2 \): \[ F\left(\frac{5\pi}{2}\right) = \sin\left(\frac{5\pi}{2}\right) - 2\left(\frac{5\pi}{2}\right) = 1 - 5\pi \] \[ F\left(-\frac{5\pi}{2}\right) = \sin\left(-\frac{5\pi}{2}\right) - 2\left(-\frac{5\pi}{2}\right) = -1 + 5\pi \] Now, we can compute the definite integral: \[ \int_{-5 \pi / 2}^{5 \pi / 2}(\cos x-2) d x = F\left(\frac{5\pi}{2}\right) - F\left(-\frac{5\pi}{2}\right) \] \[ = \left(1 - 5\pi\right) - \left(-1 + 5\pi\right) \] \[ = 1 - 5\pi + 1 - 5\pi = 2 - 10\pi \] Thus, the value of the integral is: \[ \int_{-5 \pi / 2}^{5 \pi / 2}(\cos x-2) d x = 2 - 10\pi \]