Answer
$$ - 9\cos x + 8\sin x + C$$
Work Step by Step
$$\eqalign{
& \int {\left( {9\sin x + 8\cos x} \right)} dx \cr
& {\text{use the sum rule for integration}} \cr
& = \int {9\sin x} dx + \int {8\cos x} dx \cr
& {\text{factor out the constant}} \cr
& = 9\int {\sin x} dx + 8\int {\cos x} dx \cr
& {\text{ using Basic Trigonometric integrals }}\int {\sin x} dx = - \cos x + C{\text{ and}} \cr
& \int {\cos x} dx = \sin x + C,{\text{ we obtain}} \cr
& \cr
& = 9\left( { - \cos x} \right) + 8\left( {\sin x} \right) + C \cr
& = - 9\cos x + 8\sin x + C \cr} $$