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

Options

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

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} e^x \sin x \, dx \), we will use the method of integration by parts.

### Step 1: Choose \( u \) and \( dv \)
Let:
- \( u = \sin x \) (which we will differentiate)
- \( dv = e^x \, dx \) (which we will integrate)

### Step 2: Differentiate and Integrate
Now we differentiate \( u \) and integrate \( dv \):
- \( du = \cos x \, dx \)
- \( v = e^x \)

### Step 3: Apply Integration by Parts Formula
Using the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
we substitute:
\[ I = \left[ e^x \sin x \right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} e^x \cos x \, dx \]

### Step 4: Evaluate the Boundary Terms
Now evaluate the boundary terms:
\[ \left[ e^x \sin x \right]_{0}^{\frac{\pi}{2}} = e^{\frac{\pi}{2}} \sin\left(\frac{\pi}{2}\right) - e^{0} \sin(0) = e^{\frac{\pi}{2}} \cdot 1 - 1 \cdot 0 = e^{\frac{\pi}{2}} \]
So we have:
\[ I = e^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} e^x \cos x \, dx \]

### Step 5: Define the New Integral
Let \( J = \int_{0}^{\frac{\pi}{2}} e^x \cos x \, dx \). Then we can express \( I \) in terms of \( J \):
\[ I = e^{\frac{\pi}{2}} - J \]

### Step 6: Apply Integration by Parts Again
Now we need to evaluate \( J \) using integration by parts again. Let:
- \( u = \cos x \)
- \( dv = e^x \, dx \)

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

Applying integration by parts:
\[ J = \left[ e^x \cos x \right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} e^x (-\sin x) \, dx \]
\[ J = \left[ e^x \cos x \right]_{0}^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} e^x \sin x \, dx \]

### Step 7: Evaluate the Boundary Terms for \( J \)
Evaluate the boundary terms:
\[ \left[ e^x \cos x \right]_{0}^{\frac{\pi}{2}} = e^{\frac{\pi}{2}} \cos\left(\frac{\pi}{2}\right) - e^{0} \cos(0) = e^{\frac{\pi}{2}} \cdot 0 - 1 \cdot 1 = -1 \]
Thus,
\[ J = -1 + I \]

### Step 8: Substitute \( J \) Back into the Equation for \( I \)
Now substitute \( J \) back into the equation for \( I \):
\[ I = e^{\frac{\pi}{2}} - (-1 + I) \]
\[ I = e^{\frac{\pi}{2}} + 1 - I \]
\[ 2I = e^{\frac{\pi}{2}} + 1 \]
\[ I = \frac{e^{\frac{\pi}{2}} + 1}{2} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{0}^{\frac{\pi}{2}} e^x \sin x \, dx = \frac{e^{\frac{\pi}{2}} + 1}{2} \]