\( \int_{-9 \pi / 2}^{9 \pi / 2}(\cos x-4) d x=\square \) (Type an exact answer.)

To solve the integral \( \int_{-9 \pi / 2}^{9 \pi / 2}(\cos x-4) d x \), we can split this into two parts: 1. The integral of \( \cos x \): \[ \int_{-9 \pi / 2}^{9 \pi / 2} \cos x \, dx = [\sin x]_{-9 \pi / 2}^{9 \pi / 2} \] Notice that \( \sin(-9 \pi / 2) = \sin(-\frac{9\pi}{2}) = 1 \) and \( \sin(9 \pi / 2) = \sin(\frac{9\pi}{2}) = -1 \). Therefore, \[ \sin(9 \pi / 2) - \sin(-9 \pi / 2) = -1 - 1 = -2. \] 2. The integral of \(-4\): \[ \int_{-9 \pi / 2}^{9 \pi / 2} -4 \, dx = -4 \left( x \right) \Big|_{-9 \pi / 2}^{9 \pi / 2} = -4 \left(9 \pi / 2 - (-9 \pi / 2)\right) = -4 (9 \pi) = -36 \pi. \] Now, combine both results: \[ \int_{-9 \pi / 2}^{9 \pi / 2}(\cos x - 4) \, dx = (-2) + (-36\pi) = -36\pi - 2. \] Thus, the final answer is: \[ \boxed{-36\pi - 2}. \]