`intxe^x cosx ` dx is equal to

Options

A. `e^x/2{(1-x)sinx-x cosx}`
B. `e^x/2{(1+x)sinx-x cosx}`
C. `e^x/2{(1-x)sinx+x cosx}`
D. None of these

Answer

To solve the integral \(\int x e^x \cos x \, dx\), we will use the method of integration by parts. The integration by parts formula is given by:

\[ \int u \, dv = uv - \int v \, du \]

### Step 1: Choose \(u\) and \(dv\)

Let:
- \(u = x e^x\) (which is an algebraic function)
- \(dv = \cos x \, dx\) (which is a trigonometric function)

### Step 2: Differentiate \(u\) and Integrate \(dv\)

Now we need to find \(du\) and \(v\):
- To find \(du\):
\[ du = \frac{d}{dx}(x e^x) \, dx = (e^x + x e^x) \, dx = e^x(1 + x) \, dx \]

- To find \(v\):
\[ v = \int \cos x \, dx = \sin x \]

### Step 3: Apply Integration by Parts

Now we can apply the integration by parts formula:
\[ \int x e^x \cos x \, dx = u v - \int v \, du \]
Substituting the values we found:
\[ = x e^x \sin x - \int \sin x \cdot e^x(1 + x) \, dx \]

### Step 4: Simplify the Integral

Now we need to simplify the integral:
\[ \int \sin x \cdot e^x(1 + x) \, dx = \int e^x \sin x \, dx + \int x e^x \sin x \, dx \]

### Step 5: Solve the Integral \(\int e^x \sin x \, dx\)

We will use the known formula for \(\int e^{ax} \sin(bx) \, dx\):
\[ \int e^{ax} \sin(bx) \, dx = \frac{e^{ax}}{a^2 + b^2} (a \sin(bx) - b \cos(bx)) \]
For our case, \(a = 1\) and \(b = 1\):
\[ \int e^x \sin x \, dx = \frac{e^x}{1^2 + 1^2} (1 \sin x - 1 \cos x) = \frac{e^x}{2} (\sin x - \cos x) \]

### Step 6: Substitute Back

Now substituting back into our equation:
\[ \int x e^x \cos x \, dx = x e^x \sin x - \left( \frac{e^x}{2} (\sin x - \cos x) + \int x e^x \sin x \, dx \right) \]

### Step 7: Rearranging the Equation

Rearranging gives:
\[ \int x e^x \cos x \, dx + \int x e^x \sin x \, dx = x e^x \sin x - \frac{e^x}{2} (\sin x - \cos x) \]

### Step 8: Combine the Integrals

Let \(I = \int x e^x \cos x \, dx\) and \(J = \int x e^x \sin x \, dx\):
\[ I + J = x e^x \sin x - \frac{e^x}{2} (\sin x - \cos x) \]

### Step 9: Solve for \(I\)

Now we can express \(I\) in terms of \(J\):
\[ I = x e^x \sin x - \frac{e^x}{2} (\sin x - \cos x) - J \]

### Step 10: Substitute \(J\) Back

Now we can substitute \(J\) back into our equation:
\[ I = x e^x \sin x - \frac{e^x}{2} (\sin x - \cos x) - \left( x e^x \cos x - \frac{e^x}{2} (\cos x + \sin x) \right) \]

### Final Step: Combine and Simplify

After simplifying, we get:
\[ I = \frac{e^x}{2} \left( x \sin x + x \cos x - \sin x + \cos x \right) + C \]

Thus, the final result is:
\[ \int x e^x \cos x \, dx = \frac{e^x}{2} \left( x \sin x + x \cos x - \sin x + \cos x \right) + C \]