We need to evaluate the integral:

$I = \int \cos(7x) \cos(2x) \, dx$

Step 1: Use Product-to-Sum Identity

We use the identity:

$\cos A \cos B = \frac{1}{2} [\cos(A - B) + \cos(A + B)]$

Setting $A = 7x$ and $B = 2x$, we get:

$\cos(7x) \cos(2x) = \frac{1}{2} [\cos(7x - 2x) + \cos(7x + 2x)]$

$= \frac{1}{2} [\cos(5x) + \cos(9x)]$

Step 2: Integrate Each Term

Thus, our integral becomes:

$I = \int \frac{1}{2} [\cos(5x) + \cos(9x)] \, dx$

Splitting it up:

$I = \frac{1}{2} \int \cos(5x) \, dx + \frac{1}{2} \int \cos(9x) \, dx$

We use the standard integral:

$\int \cos(ax) \, dx = \frac{\sin(ax)}{a} + C$

Applying this:

$\int \cos(5x) \, dx = \frac{\sin(5x)}{5}$

$\int \cos(9x) \, dx = \frac{\sin(9x)}{9}$

Step 3: Final Answer

$I = \frac{1}{2} \left( \frac{\sin(5x)}{5} + \frac{\sin(9x)}{9} \right) + C$

$I = \frac{\sin(5x)}{10} + \frac{\sin(9x)}{18} + C$

Conclusion:

$\int \cos(7x) \cos(2x) \, dx = \frac{\sin(5x)}{10} + \frac{\sin(9x)}{18} + C$

Would you like a step-by-step breakdown of any part?

---

Related Problems:

1. Compute $\int \cos(3x) \cos(5x) \, dx$.
2. Evaluate $\int \sin(4x) \cos(6x) \, dx$.
3. Solve $\int \cos^2(5x) \, dx$ using a trigonometric identity.
4. Find $\int \cos(2x) \sin(3x) \, dx$.
5. Derive the product-to-sum identity used in this solution.

Tip:

For trigonometric integrals involving products, using product-to-sum identities often simplifies the computation.