`inte^(-3x)cos^3x d x`

Explanation

To solve the integral \( \int e^{-3x} \cos^3 x \, dx \), we will follow these steps:

### Step 1: Use the identity for \( \cos^3 x \)
We can express \( \cos^3 x \) in terms of \( \cos x \) and \( \cos 3x \) using the identity:
\[ \cos^3 x = \frac{3 \cos x + \cos 3x}{4} \]
Thus, we rewrite the integral:
\[ \int e^{-3x} \cos^3 x \, dx = \int e^{-3x} \left( \frac{3 \cos x + \cos 3x}{4} \right) dx \]
This simplifies to:
\[ \frac{1}{4} \int e^{-3x} (3 \cos x + \cos 3x) \, dx \]

### Step 2: Split the integral
We can split the integral into two parts:
\[ \frac{1}{4} \left( 3 \int e^{-3x} \cos x \, dx + \int e^{-3x} \cos 3x \, dx \right) \]

### Step 3: Solve \( \int e^{-3x} \cos x \, dx \)
Using the formula for the integral of the form \( \int e^{ax} \cos bx \, dx \):
\[ \int e^{ax} \cos bx \, dx = \frac{e^{ax}}{a^2 + b^2} (a \cos bx + b \sin bx) \]
Here, \( a = -3 \) and \( b = 1 \):
\[ \int e^{-3x} \cos x \, dx = \frac{e^{-3x}}{(-3)^2 + 1^2} (-3 \cos x + 1 \sin x) = \frac{e^{-3x}}{9 + 1} (-3 \cos x + \sin x) = \frac{e^{-3x}}{10} (-3 \cos x + \sin x) \]

### Step 4: Solve \( \int e^{-3x} \cos 3x \, dx \)
Using the same formula, with \( a = -3 \) and \( b = 3 \):
\[ \int e^{-3x} \cos 3x \, dx = \frac{e^{-3x}}{(-3)^2 + 3^2} (-3 \cos 3x + 3 \sin 3x) = \frac{e^{-3x}}{9 + 9} (-3 \cos 3x + 3 \sin 3x) = \frac{e^{-3x}}{18} (-3 \cos 3x + 3 \sin 3x) \]

### Step 5: Combine the results
Now we substitute back into our split integral:
\[ \frac{1}{4} \left( 3 \cdot \frac{e^{-3x}}{10} (-3 \cos x + \sin x) + \frac{e^{-3x}}{18} (-3 \cos 3x + 3 \sin 3x) \right) \]

### Step 6: Simplify
Combining the terms, we have:
\[ = \frac{e^{-3x}}{4} \left( \frac{3}{10} (-3 \cos x + \sin x) + \frac{1}{18} (-3 \cos 3x + 3 \sin 3x) \right) \]

### Final Answer
Thus, the integral evaluates to:
\[ \int e^{-3x} \cos^3 x \, dx = \frac{e^{-3x}}{4} \left( \frac{3}{10} (-3 \cos x + \sin x) + \frac{1}{18} (-3 \cos 3x + 3 \sin 3x) \right) + C \]