`inte^(x)sinxcosxcos2xcos4xdx`

Options

A. `(e^(x))/(4sqrt(65))sin(8x-tan^(-1)8)+c`
B. `(e^(x))/(8sqrt(65))sin(8x-tan^(-1)8)+c`
C. `(e^(x))/(8)sin(8x-tan^(-1)8)+c`
D. none of these

Answer

To solve the integral \( \int e^x \sin x \cos x \cos 2x \cos 4x \, dx \), we can follow these steps:

### Step 1: Simplify using trigonometric identities
We know that:
\[ 2 \sin x \cos x = \sin(2x) \]
Thus, we can rewrite the integral as:
\[ \int e^x \sin x \cos x \cos 2x \cos 4x \, dx = \frac{1}{2} \int e^x \sin(2x) \cos(2x) \cos(4x) \, dx \]

### Step 2: Apply the identity again
Next, we use the identity again on \( \sin(2x) \cos(4x) \):
\[ 2 \sin(2x) \cos(4x) = \sin(6x) + \sin(2x) \]
So, we multiply and divide by 2:
\[ \frac{1}{2} \int e^x \sin(2x) \cos(4x) \, dx = \frac{1}{4} \int e^x ( \sin(6x) + \sin(2x) ) \, dx \]

### Step 3: Separate the integral
Now, we separate the integral:
\[ \frac{1}{4} \left( \int e^x \sin(6x) \, dx + \int e^x \sin(2x) \, dx \right) \]

### Step 4: Use integration by parts
We will use integration by parts for both integrals. Let’s denote:
\[ I_1 = \int e^x \sin(6x) \, dx \]
\[ I_2 = \int e^x \sin(2x) \, dx \]

For \( I_1 \):
Let \( u = e^x \) and \( dv = \sin(6x) \, dx \). Then:
\[ du = e^x \, dx, \quad v = -\frac{1}{6} \cos(6x) \]
Thus,
\[ I_1 = -\frac{1}{6} e^x \cos(6x) - \int -\frac{1}{6} \cos(6x) e^x \, dx \]
This gives:
\[ I_1 = -\frac{1}{6} e^x \cos(6x) + \frac{1}{6} I_1 \]
Solving for \( I_1 \):
\[ I_1 \left(1 - \frac{1}{6}\right) = -\frac{1}{6} e^x \cos(6x) \]
\[ I_1 \cdot \frac{5}{6} = -\frac{1}{6} e^x \cos(6x) \]
\[ I_1 = -\frac{1}{5} e^x \cos(6x) \]

For \( I_2 \):
Using the same method:
\[ I_2 = -\frac{1}{2} e^x \cos(2x) + \frac{1}{2} I_2 \]
Solving for \( I_2 \):
\[ I_2 \left(1 - \frac{1}{2}\right) = -\frac{1}{2} e^x \cos(2x) \]
\[ I_2 \cdot \frac{1}{2} = -\frac{1}{2} e^x \cos(2x) \]
\[ I_2 = -e^x \cos(2x) \]

### Step 5: Combine results
Now substituting \( I_1 \) and \( I_2 \) back into the integral:
\[ \int e^x \sin x \cos x \cos 2x \cos 4x \, dx = \frac{1}{4} \left( -\frac{1}{5} e^x \cos(6x) - e^x \cos(2x) \right) \]

### Step 6: Final result
Thus, the final result is:
\[ \int e^x \sin x \cos x \cos 2x \cos 4x \, dx = -\frac{1}{20} e^x \cos(6x) - \frac{1}{4} e^x \cos(2x) + C \]