`int e^(2x) sin x cos x dx`

Answer

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

### Step 1: Simplify the integrand using a trigonometric identity
We know that:
\[ \sin 2\theta = 2 \sin \theta \cos \theta \]
Thus, we can rewrite \( \sin x \cos x \) as:
\[ \sin x \cos x = \frac{1}{2} \sin 2x \]
Now, substituting this into the integral, we have:
\[ I = \int e^{2x} \sin x \cos x \, dx = \int e^{2x} \cdot \frac{1}{2} \sin 2x \, dx \]
This simplifies to:
\[ I = \frac{1}{2} \int e^{2x} \sin 2x \, dx \]

### Step 2: Use integration by parts or a standard formula
We can use the standard formula for the integral of the form \( \int e^{ax} \sin bx \, dx \):
\[ \int e^{ax} \sin bx \, dx = \frac{e^{ax}}{a^2 + b^2} (a \sin bx - b \cos bx) + C \]
In our case, \( a = 2 \) and \( b = 2 \).

### Step 3: Apply the formula
Substituting \( a = 2 \) and \( b = 2 \) into the formula:
\[ \int e^{2x} \sin 2x \, dx = \frac{e^{2x}}{2^2 + 2^2} (2 \sin 2x - 2 \cos 2x) + C \]
Calculating \( 2^2 + 2^2 = 4 + 4 = 8 \), we have:
\[ \int e^{2x} \sin 2x \, dx = \frac{e^{2x}}{8} (2 \sin 2x - 2 \cos 2x) + C \]

### Step 4: Substitute back into the integral
Now, substituting this back into our expression for \( I \):
\[ I = \frac{1}{2} \cdot \frac{e^{2x}}{8} (2 \sin 2x - 2 \cos 2x) + C \]
This simplifies to:
\[ I = \frac{e^{2x}}{16} (2 \sin 2x - 2 \cos 2x) + C \]

### Step 5: Final simplification
Factoring out the 2:
\[ I = \frac{e^{2x}}{16} \cdot 2 (\sin 2x - \cos 2x) + C \]
Thus:
\[ I = \frac{e^{2x}}{8} (\sin 2x - \cos 2x) + C \]

### Final Answer
\[ \int e^{2x} \sin x \cos x \, dx = \frac{e^{2x}}{8} (\sin 2x - \cos 2x) + C \]