`int_(-pi//2)^(pi//2)sin (|x|)dx=`

Options

A. 1
B. 2
C. `-1`
D. `-2`

Answer

To solve the integral \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(|x|) \, dx \), we can follow these steps:

### Step 1: Understand the function
The function \( \sin(|x|) \) means that for negative values of \( x \), the function will reflect the positive values. Therefore, we can split the integral into two parts:
\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(|x|) \, dx = \int_{-\frac{\pi}{2}}^{0} \sin(-x) \, dx + \int_{0}^{\frac{\pi}{2}} \sin(x) \, dx \]

### Step 2: Simplify the integral
Since \( \sin(-x) = -\sin(x) \), we can rewrite the first integral:
\[ \int_{-\frac{\pi}{2}}^{0} \sin(-x) \, dx = -\int_{-\frac{\pi}{2}}^{0} \sin(x) \, dx \]
Thus, we have:
\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(|x|) \, dx = -\int_{-\frac{\pi}{2}}^{0} \sin(x) \, dx + \int_{0}^{\frac{\pi}{2}} \sin(x) \, dx \]

### Step 3: Change the limits of the first integral
We can change the variable in the first integral. Let \( u = -x \), then \( du = -dx \). When \( x = -\frac{\pi}{2} \), \( u = \frac{\pi}{2} \) and when \( x = 0 \), \( u = 0 \):
\[ -\int_{-\frac{\pi}{2}}^{0} \sin(x) \, dx = \int_{0}^{\frac{\pi}{2}} \sin(u) \, du \]
So now we can combine the integrals:
\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(|x|) \, dx = \int_{0}^{\frac{\pi}{2}} \sin(x) \, dx + \int_{0}^{\frac{\pi}{2}} \sin(x) \, dx = 2 \int_{0}^{\frac{\pi}{2}} \sin(x) \, dx \]

### Step 4: Evaluate the integral
Now we need to evaluate \( \int_{0}^{\frac{\pi}{2}} \sin(x) \, dx \):
\[ \int \sin(x) \, dx = -\cos(x) + C \]
Thus,
\[ \int_{0}^{\frac{\pi}{2}} \sin(x) \, dx = \left[-\cos(x)\right]_{0}^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right) - (-\cos(0)) = 0 - (-1) = 1 \]

### Step 5: Final calculation
Now substituting back:
\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(|x|) \, dx = 2 \cdot 1 = 2 \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin(|x|) \, dx = 2 \]