`int_(0)^(pi//2) e^(x) cos x" " dx=`

Options

A. `1/2(1-e^(pi/2))`
B. `1/2(e^(pi/2)-1)`
C. `1-e^(pi/2)`
D. `e^(pi/2) -1`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} e^x \cos x \, dx \), we will use integration by parts. Let's break down the solution step by step.

### Step 1: Set Up Integration by Parts
We will use the integration by parts formula, which states:
\[ \int u \, dv = uv - \int v \, du \]
Let's choose:
- \( u = \cos x \) (which will be differentiated)
- \( dv = e^x \, dx \) (which will be integrated)

### Step 2: Differentiate and Integrate
Now we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = -\sin x \, dx \]
- Integrate \( dv \):
\[ v = e^x \]

### Step 3: Apply Integration by Parts
Now we can apply the integration by parts formula:
\[ I = \left[ e^x \cos x \right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} e^x (-\sin x) \, dx \]
This simplifies to:
\[ I = \left[ e^x \cos x \right]_{0}^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} e^x \sin x \, dx \]

### Step 4: Evaluate the Boundary Terms
Now we will evaluate \( \left[ e^x \cos x \right]_{0}^{\frac{\pi}{2}} \):
- At \( x = \frac{\pi}{2} \):
\[ e^{\frac{\pi}{2}} \cos\left(\frac{\pi}{2}\right) = e^{\frac{\pi}{2}} \cdot 0 = 0 \]
- At \( x = 0 \):
\[ e^0 \cos(0) = 1 \cdot 1 = 1 \]
Thus,
\[ \left[ e^x \cos x \right]_{0}^{\frac{\pi}{2}} = 0 - 1 = -1 \]

### Step 5: Substitute Back into the Equation
Now substituting back into our equation for \( I \):
\[ I = -1 + \int_{0}^{\frac{\pi}{2}} e^x \sin x \, dx \]

### Step 6: Apply Integration by Parts Again
Let’s denote the second integral as \( J = \int_{0}^{\frac{\pi}{2}} e^x \sin x \, dx \). We will apply integration by parts again:
- Choose:
- \( u = \sin x \)
- \( dv = e^x \, dx \)

Then:
- \( du = \cos x \, dx \)
- \( v = e^x \)

Applying integration by parts:
\[ J = \left[ e^x \sin x \right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} e^x \cos x \, dx \]
Evaluating the boundary terms:
- At \( x = \frac{\pi}{2} \):
\[ e^{\frac{\pi}{2}} \sin\left(\frac{\pi}{2}\right) = e^{\frac{\pi}{2}} \cdot 1 = e^{\frac{\pi}{2}} \]
- At \( x = 0 \):
\[ e^0 \sin(0) = 1 \cdot 0 = 0 \]
Thus,
\[ \left[ e^x \sin x \right]_{0}^{\frac{\pi}{2}} = e^{\frac{\pi}{2}} - 0 = e^{\frac{\pi}{2}} \]
So we have:
\[ J = e^{\frac{\pi}{2}} - I \]

### Step 7: Substitute \( J \) Back into the Equation for \( I \)
Now substituting \( J \) back into our equation for \( I \):
\[ I = -1 + (e^{\frac{\pi}{2}} - I) \]
This simplifies to:
\[ I + I = e^{\frac{\pi}{2}} - 1 \]
Thus,
\[ 2I = e^{\frac{\pi}{2}} - 1 \]
Finally, we find:
\[ I = \frac{e^{\frac{\pi}{2}} - 1}{2} \]

### Final Answer
\[ \int_{0}^{\frac{\pi}{2}} e^x \cos x \, dx = \frac{e^{\frac{\pi}{2}} - 1}{2} \]