`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 will follow these steps:

### Step 1: Simplify the Integral
We start by rewriting the integral:
\[ I = \int e^x \sin x \cos x \cos 2x \cos 4x \, dx \]
We can use the identity \( \sin x \cos x = \frac{1}{2} \sin 2x \):
\[ I = \int e^x \cdot \frac{1}{2} \sin 2x \cos 2x \cos 4x \, dx \]
This can be simplified further using \( \sin 2x \cos 2x = \frac{1}{2} \sin 4x \):
\[ I = \frac{1}{4} \int e^x \sin 4x \cos 4x \, dx \]

### Step 2: Simplify Further
Using the identity \( \sin 4x \cos 4x = \frac{1}{2} \sin 8x \):
\[ I = \frac{1}{8} \int e^x \sin 8x \, dx \]

### Step 3: Use Integration by Parts
Now, we apply the formula for the integral \( \int e^{ax} \sin bx \, dx \):
\[ \int e^{ax} \sin bx \, dx = \frac{e^{ax}}{a^2 + b^2} (a \sin bx - b \cos bx) + C \]
In our case, \( a = 1 \) and \( b = 8 \):
\[ \int e^x \sin 8x \, dx = \frac{e^x}{1^2 + 8^2} (1 \sin 8x - 8 \cos 8x) + C \]
Calculating \( 1^2 + 8^2 = 1 + 64 = 65 \):
\[ \int e^x \sin 8x \, dx = \frac{e^x}{65} (\sin 8x - 8 \cos 8x) + C \]

### Step 4: Substitute Back
Now substituting back into our expression for \( I \):
\[ I = \frac{1}{8} \cdot \frac{e^x}{65} (\sin 8x - 8 \cos 8x) + C \]
This simplifies to:
\[ I = \frac{e^x}{520} (\sin 8x - 8 \cos 8x) + C \]

### Step 5: Final Expression
Thus, the final expression for the integral is:
\[ I = \frac{e^x}{520} (\sin 8x - 8 \cos 8x) + C \]