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?

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.