Evaluate the following integrals.
`int sin x cos x (sin 2x + cos 2x) dx `

Answer

To evaluate the integral \( I = \int \sin x \cos x (\sin 2x + \cos 2x) \, dx \), we will follow a systematic approach.

### Step 1: Rewrite the integral
We start with the integral:
\[ I = \int \sin x \cos x (\sin 2x + \cos 2x) \, dx \]
We can use the identity \( \sin x \cos x = \frac{1}{2} \sin 2x \) to simplify the integral. Thus, we rewrite it as:
\[ I = \int \frac{1}{2} \sin 2x (\sin 2x + \cos 2x) \, dx \]

### Step 2: Distribute the terms
Now, we distribute \( \sin 2x \) inside the integral:
\[ I = \frac{1}{2} \int (\sin^2 2x + \sin 2x \cos 2x) \, dx \]

### Step 3: Break down the integral
We can split the integral into two parts:
\[ I = \frac{1}{2} \left( \int \sin^2 2x \, dx + \int \sin 2x \cos 2x \, dx \right) \]

### Step 4: Evaluate the first integral
For the first integral \( \int \sin^2 2x \, dx \), we can use the identity:
\[ \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \]
Applying this gives:
\[ \int \sin^2 2x \, dx = \int \frac{1 - \cos 4x}{2} \, dx = \frac{1}{2} \int (1 - \cos 4x) \, dx \]
Now, integrating:
\[ = \frac{1}{2} \left( x - \frac{\sin 4x}{4} \right) + C_1 \]

### Step 5: Evaluate the second integral
For the second integral \( \int \sin 2x \cos 2x \, dx \), we can use the identity \( \sin 2\theta = 2 \sin \theta \cos \theta \):
\[ \int \sin 2x \cos 2x \, dx = \frac{1}{2} \int \sin 4x \, dx \]
Integrating gives:
\[ = -\frac{1}{8} \cos 4x + C_2 \]

### Step 6: Combine results
Now, substituting back into our expression for \( I \):
\[ I = \frac{1}{2} \left( \frac{1}{2} \left( x - \frac{\sin 4x}{4} \right) + \left( -\frac{1}{8} \cos 4x \right) \right) \]
Simplifying:
\[ I = \frac{1}{4} x - \frac{1}{16} \sin 4x - \frac{1}{16} \cos 4x + C \]

### Final Answer
Thus, the final result for the integral is:
\[ I = \frac{x}{4} - \frac{\sin 4x}{16} - \frac{\cos 4x}{16} + C \]