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The given integral is:

\[
\int \sin(7x) \cos(2x) \, dx
\]

This type of integral can be simplified using a product-to-sum identity. Specifically, we use the identity:

\[
\sin(Ax)\cos(Bx) = \frac{1}{2} \left[\sin((A+B)x) + \sin((A-B)x)\right]
\]

In this case, \( A = 7 \) and \( B = 2 \), so:

\[
\sin(7x)\cos(2x) = \frac{1}{2} \left[\sin((7+2)x) + \sin((7-2)x)\right]
= \frac{1}{2} \left[\sin(9x) + \sin(5x)\right]
\]

Now the integral becomes:

\[
\int \sin(7x)\cos(2x) \, dx = \frac{1}{2} \int \left[\sin(9x) + \sin(5x)\right] dx
\]

We can now integrate each term separately:

1. \(\int \sin(9x) \, dx = -\frac{1}{9} \cos(9x)\)
2. \(\int \sin(5x) \, dx = -\frac{1}{5} \cos(5x)\)

Therefore, the result of the integral is:

\[
\frac{1}{2} \left[-\frac{1}{9} \cos(9x) - \frac{1}{5} \cos(5x)\right] + C
\]

Simplifying this:

\[
= -\frac{1}{18} \cos(9x) - \frac{1}{10} \cos(5x) + C
\]

This is the solution to the integral.

Let me know if you want further details or have any questions. Here are some related problems to explore:

1. How would you solve the integral of \(\sin(3x)\cos(4x)\)?
2. What happens when integrating products of different trigonometric functions like \(\sin(x)\cos(x)\)?
3. Can you use substitution to solve a similar integral with different arguments in sine and cosine?
4. How does the product-to-sum identity help simplify other types of integrals?
5. What techniques are available for solving integrals of trigonometric functions that are not directly product forms?

**Tip:** The product-to-sum formulas are particularly useful when dealing with integrals of products of sine and cosine functions, as they transform products into sums, making them easier to integrate.

This guide explains how to solve the integral ∫ sin(7x) cos(2x) dx by using the product-to-sum identity, a key concept in trigonometry. Detailed steps include transforming the product into a sum and then performing the integration of each term. Suitable for students in Grades 11-12 studying calculus.

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