`int_(-pi //2)^(pi//2) sin |x|dx` is equal to

Options

A. 1
B. 2
C. `-2`
D. none of these

Answer

To solve the integral \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin |x| \, dx \), we can break it down step by step.

### Step 1: Understand the function \( \sin |x| \)
The function \( \sin |x| \) is even because \( |x| \) is even. This means that \( \sin |x| = \sin x \) for \( x \geq 0 \) and \( \sin |x| = -\sin x \) for \( x < 0 \). 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: Evaluate the integral from \( -\frac{\pi}{2} \) to \( 0 \)
For \( x \) in the interval \( [-\frac{\pi}{2}, 0] \), we have \( |x| = -x \). Thus, we can rewrite the first integral:

\[ \int_{-\frac{\pi}{2}}^{0} \sin |x| \, dx = \int_{-\frac{\pi}{2}}^{0} \sin (-x) \, dx = -\int_{-\frac{\pi}{2}}^{0} \sin x \, dx \]

### Step 3: Evaluate the integral from \( 0 \) to \( \frac{\pi}{2} \)
For \( x \) in the interval \( [0, \frac{\pi}{2}] \), we have \( |x| = x \). Thus, we can rewrite the second integral:

\[ \int_{0}^{\frac{\pi}{2}} \sin |x| \, dx = \int_{0}^{\frac{\pi}{2}} \sin x \, dx \]

### Step 4: Combine the integrals
Now we can combine the 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 5: Evaluate the integrals
We know that:

\[ \int \sin x \, dx = -\cos x \]

Now we evaluate each integral:

1. For \( \int_{0}^{\frac{\pi}{2}} \sin x \, dx \):

\[ \int_{0}^{\frac{\pi}{2}} \sin x \, dx = [-\cos x]_{0}^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right) + \cos(0) = 0 + 1 = 1 \]

2. For \( \int_{-\frac{\pi}{2}}^{0} \sin x \, dx \):

\[ \int_{-\frac{\pi}{2}}^{0} \sin x \, dx = [-\cos x]_{-\frac{\pi}{2}}^{0} = -\cos(0) + \cos\left(-\frac{\pi}{2}\right) = -1 + 0 = -1 \]

### Step 6: Combine the results
Now substituting back into our combined integral:

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin |x| \, dx = -(-1) + 1 = 1 + 1 = 2 \]

### Final Answer
Thus, the value of the integral is:

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin |x| \, dx = 2 \]