Answer
$$ - 2$$
Work Step by Step
$$\eqalign{
& \int_{ - \pi /2}^{\pi /2} {15{{\sin }^4}3x\cos 3x} dx \cr
& {\text{Use the substitution method}}{\text{: }} \cr
& u = \sin 3x,\,\,\,\,du = 3\cos 3xdx,\,\,\,\,\,dx = \frac{{du}}{{3\cos x}} \cr
& {\text{Then}}{\text{,}} \cr
& \int {15{{\left( {\sin x} \right)}^4}\cos x} dx = \int {15{u^4}\cos 3x} \left( {\frac{{du}}{{3\cos 3x}}} \right) \cr
& = \int {5{u^4}du} \cr
& {\text{Integrate by the power rule}} \cr
& = 5\left( {\frac{{{u^5}}}{5}} \right) + C \cr
& = {u^5} + C \cr
& {\text{Write in terms of }}x;{\text{ replace }}u = \sin 3x \cr
& = {\left( {\sin 3x} \right)^5} + C \cr
& \int_{ - \pi /2}^{\pi /2} {15{{\sin }^4}3x\cos 3x} dx = \left( {{{\left( {\sin 3x} \right)}^5}} \right)_{ - \pi /2}^{\pi /2} \cr
& {\text{Evaluate}} \cr
& = {\left( {\sin \frac{{3\pi }}{2}} \right)^5} - {\left( {\sin \left( { - \frac{{3\pi }}{2}} \right)} \right)^5} \cr
& = {\left( { - 1} \right)^5} - {\left( 1 \right)^5} \cr
& = - 2 \cr} $$
