`int_(-pi//4)^(pi//4) e^(-x)sin x" dx"` is

Options

A. `-(sqrt(2))/(2)e^(pi//4)`
B. `(sqrt(2))/(2)e^(-pi//4)`
C. `sqrt(2)(e^(-pi//4)-e^(pi//4))`
D. zero

Answer

To solve the integral \( I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} e^{-x} \sin x \, dx \), we will use the method of integration by parts.

### Step 1: Set up the integration by parts
We choose:
- \( u = \sin x \) (which will differentiate to \( \cos x \))
- \( dv = e^{-x} \, dx \) (which will integrate to \( v = -e^{-x} \))

Using the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]

### Step 2: Apply integration by parts
Substituting our choices into the formula:
\[ I = \int e^{-x} \sin x \, dx = -e^{-x} \sin x - \int -e^{-x} \cos x \, dx \]
This simplifies to:
\[ I = -e^{-x} \sin x + \int e^{-x} \cos x \, dx \]

### Step 3: Solve the second integral using integration by parts again
Now we need to evaluate \( \int e^{-x} \cos x \, dx \). We apply integration by parts again:
- Let \( u = \cos x \) (which differentiates to \( -\sin x \))
- Let \( dv = e^{-x} \, dx \) (which integrates to \( v = -e^{-x} \))

Using the integration by parts formula again:
\[ \int e^{-x} \cos x \, dx = -e^{-x} \cos x - \int -e^{-x} (-\sin x) \, dx \]
This simplifies to:
\[ \int e^{-x} \cos x \, dx = -e^{-x} \cos x - \int e^{-x} \sin x \, dx \]

### Step 4: Substitute back into the equation for \( I \)
Now we substitute back into our equation for \( I \):
\[ I = -e^{-x} \sin x - e^{-x} \cos x - I \]
Rearranging gives:
\[ 2I = -e^{-x} (\sin x + \cos x) \]
Thus:
\[ I = -\frac{1}{2} e^{-x} (\sin x + \cos x) \]

### Step 5: Evaluate the definite integral
Now we need to evaluate this from \( -\frac{\pi}{4} \) to \( \frac{\pi}{4} \):
\[ I = -\frac{1}{2} \left[ e^{-\frac{\pi}{4}} \left( \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) \right) - e^{\frac{\pi}{4}} \left( \sin\left(-\frac{\pi}{4}\right) + \cos\left(-\frac{\pi}{4}\right) \right) \right] \]

### Step 6: Substitute the values of sine and cosine
We know that:
\[ \sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
And:
\[ \sin\left(-\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}}, \quad \cos\left(-\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
Substituting these values gives:
\[ I = -\frac{1}{2} \left[ e^{-\frac{\pi}{4}} \left( \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) - e^{\frac{\pi}{4}} \left( -\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right) \right] \]
This simplifies to:
\[ I = -\frac{1}{2} \left[ e^{-\frac{\pi}{4}} \cdot \frac{2}{\sqrt{2}} - 0 \right] \]
Thus:
\[ I = -\frac{1}{\sqrt{2}} e^{-\frac{\pi}{4}} \]

### Final Result
The value of the integral is:
\[ I = -\frac{1}{\sqrt{2}} e^{-\frac{\pi}{4}} \]