Answer
$$8\left[\frac{\sin \left(\frac{7 t}{2}\right)}{7}-\frac{\sin \left(\frac{9 t}{2}\right)}{9}\right]+C$$
Work Step by Step
Use the formula
$$\int \sin a x \sin b x d x=\frac{\sin (a-b) x}{2(a-b)}-\frac{\sin (a+b) x}{2(a+b)}+C, \quad a^{2} \neq b^{2}$$
with $a=4 , \ \ b=\dfrac{1}{2} $, we get
\begin{align*}
\int 8 \sin 4 t \sin \frac{t}{2} d x&=\frac{8}{7} \sin \left(\frac{7 t}{2}\right)-\frac{8}{9} \sin \left(\frac{9 t}{2}\right)+C\\
&=8\left[\frac{\sin \left(\frac{7 t}{2}\right)}{7}-\frac{\sin \left(\frac{9 t}{2}\right)}{9}\right]+C
\end{align*}
