## Calculus: Early Transcendentals 8th Edition

$$\int\sin2\theta\sin6\theta d\theta=\frac{1}{8}\sin4\theta-\frac{1}{16}\sin8\theta+C$$
$$A=\int\sin2\theta\sin6\theta d\theta$$ For this problem, we need to utilize $$\sin A\sin B=\frac{1}{2}[\cos(A-B)-\cos(A+B)]$$ Therefore, $$A=\frac{1}{2}\int[\cos(2\theta-6\theta)-\cos(2\theta+6\theta)]d\theta$$ $$A=\frac{1}{2}\int[\cos(-4\theta)-\cos8\theta]d\theta$$ $$A=\frac{1}{2}\int[\cos(4\theta)-\cos8\theta]d\theta$$ (since $\cos(-X)=\cos X$) $$A=\frac{1}{2}(\int\cos4\theta d\theta-\int\cos8\theta d\theta)$$ $$A=\frac{1}{2}[(\frac{1}{4}\int\cos4\theta d(4\theta)-\frac{1}{8}\int\cos8\theta d(8\theta)]$$ $$A=\frac{1}{2}[\frac{1}{4}\sin4\theta-\frac{1}{8}\sin8\theta]+C$$ $$A=\frac{1}{8}\sin4\theta-\frac{1}{16}\sin8\theta+C$$