`int" cos"^(3)(3x +5) dx`

Answer

To solve the integral \( \int \cos^3(3x + 5) \, dx \), we will follow a systematic approach using substitution and integration techniques. Here’s the step-by-step solution:

### Step 1: Substitution
Let \( t = 3x + 5 \). Then, differentiate both sides with respect to \( x \):
\[ \frac{dt}{dx} = 3 \implies dx = \frac{dt}{3} \]

### Step 2: Rewrite the Integral
Substituting \( t \) into the integral, we have:
\[ \int \cos^3(3x + 5) \, dx = \int \cos^3(t) \cdot \frac{dt}{3} = \frac{1}{3} \int \cos^3(t) \, dt \]

### Step 3: Use Trigonometric Identity
We can rewrite \( \cos^3(t) \) using the identity:
\[ \cos^3(t) = \cos(t) \cdot \cos^2(t) = \cos(t) \cdot (1 - \sin^2(t)) \]
Thus, the integral becomes:
\[ \frac{1}{3} \int \cos(t) (1 - \sin^2(t)) \, dt = \frac{1}{3} \left( \int \cos(t) \, dt - \int \cos(t) \sin^2(t) \, dt \right) \]

### Step 4: First Integral
The first integral is straightforward:
\[ \int \cos(t) \, dt = \sin(t) \]

### Step 5: Second Integral
For the second integral, we use substitution again. Let \( u = \sin(t) \), then \( du = \cos(t) \, dt \):
\[ \int \cos(t) \sin^2(t) \, dt = \int u^2 \, du = \frac{u^3}{3} = \frac{\sin^3(t)}{3} \]

### Step 6: Combine Results
Now, substituting back into our integral:
\[ \frac{1}{3} \left( \sin(t) - \frac{\sin^3(t)}{3} \right) = \frac{1}{3} \sin(t) - \frac{1}{9} \sin^3(t) \]

### Step 7: Substitute Back
Now we substitute back \( t = 3x + 5 \):
\[ = \frac{1}{3} \sin(3x + 5) - \frac{1}{9} \sin^3(3x + 5) + C \]

### Final Result
Thus, the final result of the integral is:
\[ \int \cos^3(3x + 5) \, dx = \frac{1}{3} \sin(3x + 5) - \frac{1}{9} \sin^3(3x + 5) + C \]