Answer
$$\int \sin 5 x \sin 4 x d x =\frac{1}{2} \sin x-\frac{1}{18} \sin (9x)+c$$
Work Step by Step
$$
\int \sin 5 x \sin 4 x d x
$$
Since $$
\sin m x \sin n x=\frac{1}{2}(\cos [(m-n) x]-\cos [(m+n) x])
$$
Then
\begin{align*}
\int \sin 5 x \sin 4 x d x&=\frac{1}{2}\int \cos (x)- \cos (9 x) d x\\
&=\frac{1}{2} \sin x-\frac{1}{18} \sin (9x)+c
\end{align*}
