Answer
$$\int\cos\frac{\theta}{2}\cos7\theta d\theta=\frac{\sin\frac{13\theta}{2}}{13}+\frac{\sin\frac{15\theta}{2}}{15}+C$$
Work Step by Step
$$A=\int\cos\frac{\theta}{2}\cos7\theta d\theta$$
Use Formula 69c, which states that
$$\int \cos ax\cos bx dx=\frac{\sin(a-b)x}{2(a-b)}+\frac{\sin(a+b)x}{2(a+b)}+C$$
for $a=1/2$ and $b=7$:
$a-b=1/2-7=-13/2$ while $a+b=1/2+7=15/2$
$$A=\frac{\sin\Big(-\frac{13}{2}\Big)\theta}{2\times\Big(-\frac{13}{2}\Big)}+\frac{\sin\frac{15}{2}\theta}{2\times\frac{15}{2}}+C$$
We have $\sin\Big(-\frac{13}{2}\Big)\theta=-\sin\frac{13}{2}\theta$
$$A=\frac{-\sin\frac{13\theta}{2}}{-13}+\frac{\sin\frac{15\theta}{2}}{15}+C$$ $$A=\frac{\sin\frac{13\theta}{2}}{13}+\frac{\sin\frac{15\theta}{2}}{15}+C$$
