What is `int_(-pi//2)^(pi//2) |sinx| dx` equal to ?

Options

A. `2`
B. `1`
C. `pi`
D. `0`

Explanation

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

### Step 1: Understand the function \( |\sin x| \)
The function \( \sin x \) is negative in the interval \( \left[-\frac{\pi}{2}, 0\right] \) and positive in the interval \( \left[0, \frac{\pi}{2}\right] \). Therefore, we can express \( |\sin x| \) as:
\[ |\sin x| = \begin{cases} -\sin x & \text{for } x \in \left[-\frac{\pi}{2}, 0\right] \\ \sin x & \text{for } x \in \left[0, \frac{\pi}{2}\right] \end{cases} \]

### Step 2: Break the integral into two parts
We can split the integral into two parts based on the intervals where \( \sin x \) is negative and positive:
\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} |\sin x| \, dx = \int_{-\frac{\pi}{2}}^{0} -\sin x \, dx + \int_{0}^{\frac{\pi}{2}} \sin x \, dx \]

### Step 3: Calculate the first integral
Now we calculate the first integral:
\[ \int_{-\frac{\pi}{2}}^{0} -\sin x \, dx = -\left[-\cos x\right]_{-\frac{\pi}{2}}^{0} = -\left(-\cos(0) + \cos\left(-\frac{\pi}{2}\right)\right) \]
Evaluating the limits:
\[ = -\left(-1 + 0\right) = 1 \]

### Step 4: Calculate the second integral
Next, we calculate the second integral:
\[ \int_{0}^{\frac{\pi}{2}} \sin x \, dx = \left[-\cos x\right]_{0}^{\frac{\pi}{2}} = -\left(\cos\left(\frac{\pi}{2}\right) - \cos(0)\right) \]
Evaluating the limits:
\[ = -\left(0 - 1\right) = 1 \]

### Step 5: Combine the results
Now, we combine the results of both integrals:
\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} |\sin x| \, dx = 1 + 1 = 2 \]

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

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