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The problem in the image asks to evaluate the integral:

\[
\int \sin(5x) \cos(2x) \, dx
\]

To solve this, we can use a trigonometric identity to simplify the product of sine and cosine. Specifically, the identity for the product of sine and cosine is:

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

Here, \( A = 5x \) and \( B = 2x \). Applying the identity:

\[
\sin(5x) \cos(2x) = \frac{1}{2} \left[ \sin(5x + 2x) + \sin(5x - 2x) \right]
= \frac{1}{2} \left[ \sin(7x) + \sin(3x) \right]
\]

Now, the integral becomes:

\[
\int \sin(5x) \cos(2x) \, dx = \frac{1}{2} \int \left[ \sin(7x) + \sin(3x) \right] dx
\]

We can now integrate each term:

\[
\int \sin(7x) \, dx = -\frac{1}{7} \cos(7x)
\]
\[
\int \sin(3x) \, dx = -\frac{1}{3} \cos(3x)
\]

Thus, the solution is:

\[
\frac{1}{2} \left[ -\frac{1}{7} \cos(7x) - \frac{1}{3} \cos(3x) \right] + C
\]

Simplifying:

\[
-\frac{1}{14} \cos(7x) - \frac{1}{6} \cos(3x) + C
\]

This is the evaluated integral.

Would you like a deeper explanation on any part of this? Here are some related questions:

1. How do we derive the identity \(\sin(A)\cos(B)\)?
2. Can you use the same method for other trigonometric integrals?
3. What are the conditions under which this method applies?
4. How do you apply the chain rule in reverse to integrate sine and cosine functions?
5. How would this problem change if the integral limits were given?

**Tip:** Always remember trigonometric identities—they can greatly simplify integrals involving products of sine and cosine!

This page walks through the step-by-step process of evaluating the integral of sin(5x) cos(2x) using trigonometric identities. Learn how to simplify the integral with the identity sin(A) cos(B) and integrate trigonometric functions. Suitable for high school students in Grades 10-12.

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