Evaluate the following integrals:
`int_(-pi/4)^(pi/4)|sin x| dx`

Answer

To evaluate the integral \( \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} |\sin x| \, dx \), we can follow these steps:

### Step 1: Analyze the function \( |\sin x| \)
The sine function \( \sin x \) is positive in the interval \( \left[0, \frac{\pi}{4}\right] \) and negative in the interval \( \left[-\frac{\pi}{4}, 0\right] \). Therefore, we can express the absolute value function as:
\[ |\sin x| = \begin{cases} -\sin x & \text{if } x \in \left[-\frac{\pi}{4}, 0\right] \\ \sin x & \text{if } x \in \left[0, \frac{\pi}{4}\right] \end{cases} \]

### Step 2: Split the integral
We can split the integral into two parts:
\[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} |\sin x| \, dx = \int_{-\frac{\pi}{4}}^{0} -\sin x \, dx + \int_{0}^{\frac{\pi}{4}} \sin x \, dx \]

### Step 3: Evaluate the first integral
Now, we evaluate the first integral:
\[ \int_{-\frac{\pi}{4}}^{0} -\sin x \, dx = -\left[-\cos x\right]_{-\frac{\pi}{4}}^{0} = \left[\cos x\right]_{-\frac{\pi}{4}}^{0} \]
Calculating the values:
\[ = \cos(0) - \cos\left(-\frac{\pi}{4}\right) = 1 - \frac{1}{\sqrt{2}} = 1 - \frac{\sqrt{2}}{2} \]

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

### Step 5: Combine the results
Now, we combine the results of both integrals:
\[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} |\sin x| \, dx = \left(1 - \frac{\sqrt{2}}{2}\right) + \left(1 - \frac{\sqrt{2}}{2}\right) = 2\left(1 - \frac{\sqrt{2}}{2}\right) \]
Simplifying this gives:
\[ = 2 - \sqrt{2} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} |\sin x| \, dx = 2 - \sqrt{2} \]