Find: `int cos 6x sqrt(1+sin 6x) dx`.

Answer

To solve the integral \( \int \cos(6x) \sqrt{1 + \sin(6x)} \, dx \), we can use a substitution method. Here are the steps:

### Step 1: Substitution
Let \( t = 1 + \sin(6x) \). Then, we differentiate \( t \) with respect to \( x \):
\[ \frac{dt}{dx} = 6 \cos(6x) \]
This implies:
\[ dt = 6 \cos(6x) \, dx \quad \Rightarrow \quad \cos(6x) \, dx = \frac{1}{6} dt \]

### Step 2: Rewrite the Integral
Now, substitute \( t \) into the integral:
\[ \int \cos(6x) \sqrt{1 + \sin(6x)} \, dx = \int \sqrt{t} \cdot \frac{1}{6} dt = \frac{1}{6} \int \sqrt{t} \, dt \]

### Step 3: Integrate
The integral of \( \sqrt{t} \) can be computed as follows:
\[ \int \sqrt{t} \, dt = \int t^{1/2} \, dt = \frac{t^{3/2}}{3/2} = \frac{2}{3} t^{3/2} \]
Thus,
\[ \frac{1}{6} \int \sqrt{t} \, dt = \frac{1}{6} \cdot \frac{2}{3} t^{3/2} = \frac{1}{9} t^{3/2} \]

### Step 4: Substitute Back
Now, substitute back \( t = 1 + \sin(6x) \):
\[ \frac{1}{9} t^{3/2} = \frac{1}{9} (1 + \sin(6x))^{3/2} \]

### Step 5: Add the Constant of Integration
Finally, we add the constant of integration \( C \):
\[ \int \cos(6x) \sqrt{1 + \sin(6x)} \, dx = \frac{1}{9} (1 + \sin(6x))^{3/2} + C \]

### Final Answer
Thus, the solution to the integral is:
\[ \int \cos(6x) \sqrt{1 + \sin(6x)} \, dx = \frac{1}{9} (1 + \sin(6x))^{3/2} + C \]