If `int cos x cos 2x cos 3x dx`= `A_(1)sin 2x + A_(2) sin 4x + A_(3) sin 6x + C` then

Options

A. `A_(1)=(1)/(2),A_(2)=(1)/(4)`
B. `A_(1) = (1)/(4),A_(2) = (1)/(8)`
C. `A_(2)=(1)/(16),A_(3)=(1)/(8)`
D. `A_(1)=(1)/(8),A_(3)=(1)/(24)`

Answer

To solve the integral \( I = \int \cos x \cos 2x \cos 3x \, dx \), we can use trigonometric identities to simplify the expression. Let's go through the steps systematically.

### Step 1: Use the Product-to-Sum Formulas
We start by using the product-to-sum identities. The formula for the product of cosines is:
\[ 2 \cos A \cos B = \cos(A + B) + \cos(A - B) \]
We can apply this to \( \cos x \cos 3x \):
\[ \cos x \cos 3x = \frac{1}{2} \left( \cos(4x) + \cos(-2x) \right) = \frac{1}{2} \left( \cos(4x) + \cos(2x) \right) \]
Thus, we can rewrite the integral as:
\[ I = \int \cos 2x \cdot \frac{1}{2} \left( \cos(4x) + \cos(2x) \right) \, dx \]

### Step 2: Distribute the Integral
Now, we can distribute the integral:
\[ I = \frac{1}{2} \int \cos 2x \cos 4x \, dx + \frac{1}{2} \int \cos^2 2x \, dx \]

### Step 3: Solve Each Integral
1. **Integral of \( \cos 2x \cos 4x \)**:
Using the product-to-sum formula again:
\[ 2 \cos 2x \cos 4x = \cos(6x) + \cos(-2x) = \cos(6x) + \cos(2x) \]
Therefore,
\[ \int \cos 2x \cos 4x \, dx = \frac{1}{2} \left( \int \cos(6x) \, dx + \int \cos(2x) \, dx \right) \]
The integrals are:
\[ \int \cos(6x) \, dx = \frac{1}{6} \sin(6x) + C_1 \]
\[ \int \cos(2x) \, dx = \frac{1}{2} \sin(2x) + C_2 \]
Thus,
\[ \int \cos 2x \cos 4x \, dx = \frac{1}{2} \left( \frac{1}{6} \sin(6x) + \frac{1}{2} \sin(2x) \right) = \frac{1}{12} \sin(6x) + \frac{1}{4} \sin(2x) \]

2. **Integral of \( \cos^2 2x \)**:
Using the identity \( \cos^2 A = \frac{1 + \cos(2A)}{2} \):
\[ \int \cos^2 2x \, dx = \int \frac{1 + \cos(4x)}{2} \, dx = \frac{1}{2} \left( x + \frac{1}{4} \sin(4x) \right) \]

### Step 4: Combine the Results
Now we can combine the results of the two integrals:
\[ I = \frac{1}{2} \left( \frac{1}{12} \sin(6x) + \frac{1}{4} \sin(2x) \right) + \frac{1}{4} \left( x + \frac{1}{4} \sin(4x) \right) \]
This simplifies to:
\[ I = \frac{1}{24} \sin(6x) + \frac{1}{8} \sin(2x) + \frac{1}{8} x + \frac{1}{16} \sin(4x) + C \]

### Step 5: Identify Coefficients
From the expression \( I = A_1 \sin(2x) + A_2 \sin(4x) + A_3 \sin(6x) + C \), we can identify:
- \( A_1 = \frac{1}{8} \)
- \( A_2 = \frac{1}{16} \)
- \( A_3 = \frac{1}{24} \)

### Final Answer
Thus, the values of \( A_1, A_2, A_3 \) are:
- \( A_1 = \frac{1}{8} \)
- \( A_2 = \frac{1}{16} \)
- \( A_3 = \frac{1}{24} \)