Answer
$$\pi $$
Work Step by Step
$$\eqalign{
& \int_{ - \pi }^\pi {\sin 3x} \sin 3xdx \cr
& {\text{Use the identity }}\sin mx\sin xnx = \frac{1}{2}\left[ {\cos \left( {m - n} \right)x - \cos \left( {m + n} \right)x} \right]\,\,\left( {{\text{see page 463}}} \right) \cr
& \int_{ - \pi }^\pi {\sin 3x} \sin 3xdx = \int_{ - \pi }^\pi {\frac{1}{2}\left[ {\cos \left( {3 - 3} \right)x - \cos \left( {3 + 3} \right)x} \right]dx} \cr
& = \int_{ - \pi }^\pi {\frac{1}{2}\left[ {1 - \cos 6x} \right]dx} \cr
& = \frac{1}{2}\int_{ - \pi }^\pi {\left( {1 - \cos 6x} \right)dx} \cr
& {\text{integrating}} \cr
& = \frac{1}{2}\left( {x - \frac{1}{6}\sin 6x} \right)_{ - \pi }^\pi \cr
& = \frac{1}{2}\left( {\pi - \frac{1}{6}\sin 6\pi } \right) - \frac{1}{2}\left( { - \pi - \frac{1}{6}\sin 6\left( { - \pi } \right)} \right) \cr
& {\text{simplifying}} \cr
& = \frac{1}{2}\left( {\pi - 0} \right) - \frac{1}{2}\left( { - \pi - 0} \right) \cr
& = \frac{1}{2}\pi + \frac{1}{2}\pi \cr
& = \pi \cr} $$
