Answer
$$ 0$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( x \right) = \cos 2x + \cos x\sin x - 3\sin {x^5} \cr
& {\text{Evaluate }}f\left( { - x} \right) \cr
& f\left( { - x} \right) = \cos 2\left( { - x} \right) + \cos \left( { - x} \right)\sin \left( { - x} \right) - 3\sin {\left( { - x} \right)^5} \cr
& f\left( { - x} \right) = \cos \left( { - 2x} \right) + \cos \left( { - x} \right)\sin \left( { - x} \right) - 3\sin \left( { - {x^5}} \right) \cr
& {\text{Recall that }}\cos \left( { - \theta } \right) = \cos \theta ,\,\,\,\,\sin \left( { - \theta } \right) = - \sin \theta \cr
& f\left( { - x} \right) = \cos 2x - \cos x\sin x + 3\sin {x^5} \cr
& {\text{The function is neither odd nor even so the Theorem 5}}{\text{.4 cannot}} \cr
& {\text{be applied directly}}{\text{.}} \cr
& {\text{However, we can split the integrand and then use symmetry:}} \cr
& \int_{ - \pi /2}^{\pi /2} {\left( {\cos 2x + \cos x\sin x - 3\sin {x^5}} \right)} dx \cr
& = \int_{ - \pi /2}^{\pi /2} {\cos 2x} dx + \int_{ - \pi /2}^{\pi /2} {\cos x\sin x} dx - 3\int_{ - \pi /2}^{\pi /2} {\sin {x^5}dx} \cr
& {\text{Where}} \cr
& \cos 2x:{\text{ even}} \cr
& \cos x\sin x:{\text{ odd}} \cr
& - 3\sin {x^5}:{\text{ odd}} \cr
& {\text{Therefore,}} \cr
& = 2\int_0^{\pi /2} {\cos 2x} dx + 0 - 0 \cr
& {\text{Integrating}} \cr
& = \left[ {\sin 2x} \right]_0^{\pi /2} \cr
& = \sin 2\left( {\frac{\pi }{2}} \right) - \sin \left( 0 \right) \cr
& = 0 \cr} $$
