Question

Find the integral. $$ \int \cos 4 \theta \cos (-3 \theta) d \theta $$

Short answer

The integration results in \(\frac{1}{14}\sin(7\theta) + \frac{1}{2}\sin(\theta) + C\).

Step-by-step solution

Apply the Product-to-Sum Identity

To simplify the integration process, first, apply the product-to-sum identity to the integrand. The identity for products of cosine functions is \(\cos A \cos B = \frac{1}{2} [\cos(A - B) + \cos(A + B)]\). Applying this identity gives: \[\int \cos 4 \theta \cos (-3 \theta) d \theta = \frac{1}{2} \int [\cos(4\theta - -3\theta) + \cos(4\theta + -3\theta)] d\theta\] This simplifies to: \[\frac{1}{2} \int [\cos(7\theta) + \cos(\theta)] d\theta\]

Integration

Now that the integrand is simplified, execute the integration. As you likely know, the integral of \(\cos A\theta\) is \(\frac{1}{A} \sin A\theta\). So completing the integration gives: \[\frac{1}{2} [\frac{1}{7} \sin(7\theta) + \sin(\theta) ] + C\]

Simplify the Result

Simplify the final result by distributing the 1/2: \[= \frac{1}{14}\sin(7\theta) + \frac{1}{2}\sin(\theta) + C\]