$$ \int ( 1 + \sin ^ { 2 } 2 t ) \cos ^ { 3 } 2 t . d t $$

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To solve the integral

$$
\int (1 + \sin^2(2t)) \cos^3(2t) \, dt,
$$

we can break it into two separate integrals:

$$
\int (1 + \sin^2(2t)) \cos^3(2t) \, dt = \int \cos^3(2t) \, dt + \int \sin^2(2t) \cos^3(2t) \, dt.
$$

### Step 1: Solve $$ \int \cos^3(2t) \, dt $$

We can use the identity $$ \cos^3(x) = \cos(x)(1 - \sin^2(x)) $$:

$$
\int \cos^3(2t) \, dt = \int \cos(2t)(1 - \sin^2(2t)) \, dt.
$$

Let $$ u = \sin(2t) $$, then $$ du = 2\cos(2t) \, dt $$ or $$ dt = \frac{du}{2\cos(2t)} $$. Thus, we have:

$$
\int \cos^3(2t) \, dt = \frac{1}{2} \int (1 - u^2) \, du = \frac{1}{2} \left( u - \frac{u^3}{3} \right) + C = \frac{1}{2} \left( \sin(2t) - \frac{\sin^3(2t)}{3} \right) + C.
$$

### Step 2: Solve $$ \int \sin^2(2t) \cos^3(2t) \, dt $$

Using the substitution $$ u = \sin(2t) $$ again, we have:

$$
\sin^2(2t) = u^2 \quad \text{and} \quad \cos(2t) = \sqrt{1 - u^2}.
$$

Thus,

$$
\cos^3(2t) = (1 - u^2)^{3/2}.
$$

The integral becomes:

$$
\int \sin^2(2t) \cos^3(2t) \, dt = \int u^2 (1 - u^2)^{3/2} \frac{du}{2\sqrt{1 - u^2}} = \frac{1}{2} \int u^2 (1 - u^2) \, du.
$$

Now, we can expand and integrate:

$$
\frac{1}{2} \int (u^2 - u^4) \, du = \frac{1}{2} \left( \frac{u^3}{3} - \frac{u^5}{5} \right) + C = \frac{1}{6} u^3 - \frac{1}{10} u^5 + C.
$$

Substituting back $$ u = \sin(2t) $$:

$$
\frac{1}{6} \sin^3(2t) - \frac{1}{10} \sin^5(2t) + C.
$$

### Final Result

Combining both parts, we have:

$$
\int (1 + \sin^2(2t)) \cos^3(2t) \, dt = \frac{1}{2} \left( \sin(2t) - \frac{\sin^3(2t)}{3} \right) + \frac{1}{6} \sin^3(2t) - \frac{1}{10} \sin^5(2t) + C.
$$

This simplifies to:

$$
\int (1 + \sin^2(2t)) \cos^3(2t) \, dt = \frac{1}{2} \sin(2t) + \left( \frac{1}{6} - \frac{1}{6} \right) \sin^3(2t) - \frac{1}{10} \sin^5(2t) + C.
$$

Thus, the final answer is:

$$
\int (1 + \sin^2(2t)) \cos^3(2t) \, dt = \frac{1}{2} \sin(2t) - \frac{1}{10} \sin^5(2t) + C.
$$
$$ \int (1 + \sin^2(2t)) \cos^3(2t) \, dt = \frac{1}{2} \sin(2t) - \frac{1}{10} \sin^5(2t) + C. $$

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