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 simplify and integrate step by step.

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

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

### Step 3: Use identities to simplify further
We can use the identity for \( \sin^2 2x \):
\[ \sin^2 2x = \frac{1 - \cos 4x}{2} \]
And for \( \sin 2x \cos 2x \):
\[ \sin 2x \cos 2x = \frac{1}{2} \sin 4x \]
Substituting these identities into the integral gives:
\[ I = \frac{1}{2} \int \left( \frac{1 - \cos 4x}{2} + \frac{1}{2} \sin 4x \right) \, dx \]
This simplifies to:
\[ I = \frac{1}{4} \int (1 - \cos 4x + \sin 4x) \, dx \]

### Step 4: Integrate each term
Now we can integrate term by term:
\[ I = \frac{1}{4} \left( \int 1 \, dx - \int \cos 4x \, dx + \int \sin 4x \, dx \right) \]
Calculating each integral:
1. \( \int 1 \, dx = x \)
2. \( \int \cos 4x \, dx = \frac{1}{4} \sin 4x \)
3. \( \int \sin 4x \, dx = -\frac{1}{4} \cos 4x \)

Putting it all together:
\[ I = \frac{1}{4} \left( x - \frac{1}{4} \sin 4x - \frac{1}{4} \cos 4x \right) + C \]
This simplifies to:
\[ I = \frac{x}{4} - \frac{1}{16} \sin 4x + \frac{1}{16} \cos 4x + C \]

### Final Answer
Thus, the evaluated integral is:
\[ I = \frac{x}{4} - \frac{1}{16} \sin 4x + \frac{1}{16} \cos 4x + C \]