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To evaluate the integral

\[
\int \sin^2(2x) \, dx
\]

we can use the trigonometric identity to simplify \(\sin^2(2x)\). The identity is:

\[
\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}
\]

Applying this identity with \(\theta = 2x\):

\[
\sin^2(2x) = \frac{1 - \cos(4x)}{2}
\]

Now the integral becomes:

\[
\int \sin^2(2x) \, dx = \int \frac{1 - \cos(4x)}{2} \, dx
\]

This can be split into two integrals:

\[
= \frac{1}{2} \int 1 \, dx - \frac{1}{2} \int \cos(4x) \, dx
\]

Evaluating each part separately:

1. For \(\frac{1}{2} \int 1 \, dx\):

   \[
   \frac{1}{2} \int 1 \, dx = \frac{1}{2} x
   \]

2. For \(-\frac{1}{2} \int \cos(4x) \, dx\):

   We need to integrate \(\cos(4x)\), which gives:

   \[
   \int \cos(4x) \, dx = \frac{\sin(4x)}{4}
   \]

   Therefore:

   \[
   -\frac{1}{2} \int \cos(4x) \, dx = -\frac{1}{2} \cdot \frac{\sin(4x)}{4} = -\frac{\sin(4x)}{8}
   \]

Combining these results, we get:

\[
\int \sin^2(2x) \, dx = \frac{1}{2} x - \frac{\sin(4x)}{8} + C
\]

where \(C\) is the constant of integration.

Would you like further explanation on any steps?

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Here are some related questions for further exploration:

1. How would the solution change if the integrand were \(\sin^2(3x)\) instead?
2. Can you evaluate \(\int \cos^2(2x) \, dx\) using a similar approach?
3. How does the identity \(\sin^2(x) = \frac{1 - \cos(2x)}{2}\) help simplify integrals of trigonometric squares?
4. What happens if you use a different trigonometric identity for \(\sin^2(2x)\)?
5. How would this integral look if we had \(\int \sin^2(2x) e^x \, dx\)?

**Tip:** Always consider using trigonometric identities for powers of sine and cosine; they often simplify integrals substantially.

Evaluate the integral \( \int \sin^2(2x) \, dx \).

Learn how to evaluate the integral \( \int \sin^2(2x) \, dx \) by using trigonometric identities. This solution covers key concepts in trigonometric integration and power reduction identities, suitable for high school students in Grades 11-12.

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