`"If " int x e^(x) cosx dx=ae^(x)(b(1-x)sinx+cx cosx)+d,` then

Options

A. `a=1,b=1,c=-1`
B. `a=(1)/(2),b=-1,c=1`
C. `a=1,b=-1,c=1`
D. `a=(1)/(2),b=1,c=-1`

Explanation

To solve the integral \( \int x e^x \cos x \, dx \), we will use integration by parts and the technique of reduction formulas.

### Step-by-Step Solution:

1. **Identify the parts for integration by parts**:
We will let:
\[ u = x \quad \text{and} \quad dv = e^x \cos x \, dx \]
Then, we differentiate \( u \) and integrate \( dv \):
\[ du = dx \quad \text{and} \quad v = \int e^x \cos x \, dx \]

2. **Calculate \( v \)**:
To find \( v \), we need to compute \( \int e^x \cos x \, dx \). We will apply integration by parts again:
Let:
\[ w = e^x \quad \text{and} \quad dz = \cos x \, dx \]
Then:
\[ dw = e^x \, dx \quad \text{and} \quad z = \sin x \]
Using integration by parts:
\[ \int e^x \cos x \, dx = e^x \sin x - \int e^x \sin x \, dx \]

3. **Calculate \( \int e^x \sin x \, dx \)**:
Again, we apply integration by parts:
Let:
\[ w = e^x \quad \text{and} \quad dz = \sin x \, dx \]
Then:
\[ dw = e^x \, dx \quad \text{and} \quad z = -\cos x \]
Thus:
\[ \int e^x \sin x \, dx = -e^x \cos x - \int -e^x \cos x \, dx \]
Simplifying gives:
\[ \int e^x \sin x \, dx = -e^x \cos x + \int e^x \cos x \, dx \]

4. **Combine the results**:
Let \( I = \int e^x \cos x \, dx \) and \( J = \int e^x \sin x \, dx \). We have:
\[ I = e^x \sin x - J \]
And:
\[ J = -e^x \cos x + I \]
Substituting \( J \) into the equation for \( I \):
\[ I = e^x \sin x - (-e^x \cos x + I) \]
Rearranging gives:
\[ I + I = e^x \sin x + e^x \cos x \]
\[ 2I = e^x (\sin x + \cos x) \]
Thus:
\[ I = \frac{1}{2} e^x (\sin x + \cos x) \]

5. **Substituting back into the original integral**:
Now substituting \( I \) back into our integration by parts formula:
\[ \int x e^x \cos x \, dx = x \cdot \frac{1}{2} e^x (\sin x + \cos x) - \int \frac{1}{2} e^x (\sin x + \cos x) \, dx \]
The second integral can be computed as:
\[ \int e^x (\sin x + \cos x) \, dx = \frac{1}{2} e^x (\sin x + \cos x) \]
Thus:
\[ \int x e^x \cos x \, dx = \frac{1}{2} x e^x (\sin x + \cos x) - \frac{1}{2} \cdot \frac{1}{2} e^x (\sin x + \cos x) + C \]

6. **Final result**:
Thus, we have:
\[ \int x e^x \cos x \, dx = \frac{1}{2} e^x \left( x (\sin x + \cos x) - \frac{1}{2} (\sin x + \cos x) \right) + C \]