Answer
$$\int8\sin4t\sin\frac{t}{2}dt=\frac{8}{7}\sin\frac{7}{2}t-\frac{8}{9}\sin\frac{9}{2}t+C$$
Work Step by Step
$$A=\int8\sin4t\sin\frac{t}{2}dt$$
Use Formula 69b, which states that
$$\int \sin ax\sin bx dx=\frac{\sin(a-b)x}{2(a-b)}-\frac{\sin(a+b)x}{2(a+b)}+C$$
for $a=4$ and $b=1/2$:
$$A=8\Big(\frac{\sin\frac{7}{2}t}{2\times\frac{7}{2}}-\frac{\sin\frac{9}{2}t}{2\times\frac{9}{2}}\Big)+C$$ $$A=8\Big(\frac{\sin\frac{7}{2}t}{7}-\frac{\sin\frac{9}{2}t}{9}\Big)+C$$ $$A=\frac{8}{7}\sin\frac{7}{2}t-\frac{8}{9}\sin\frac{9}{2}t+C$$
