The correct evaluation of `int_0^(pi//2)|sin(x-pi/4)|dx` is

Options

A. `2+sqrt2`
B. `2-sqrt2`
C. `-2+sqrt2`
D. 0

Answer

To evaluate the integral \( I = \int_0^{\frac{\pi}{2}} |\sin(x - \frac{\pi}{4})| \, dx \), we need to analyze the expression inside the absolute value.

### Step 1: Determine where \( \sin(x - \frac{\pi}{4}) \) changes sign
The sine function changes sign at \( x - \frac{\pi}{4} = 0 \), which gives us \( x = \frac{\pi}{4} \). Therefore, we will split the integral at this point.

### Step 2: Split the integral
We can express the integral as:
\[ I = \int_0^{\frac{\pi}{4}} -\sin(x - \frac{\pi}{4}) \, dx + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin(x - \frac{\pi}{4}) \, dx \]

### Step 3: Evaluate the first integral
For \( x \) in the interval \( [0, \frac{\pi}{4}] \), \( \sin(x - \frac{\pi}{4}) \) is negative. Thus, we have:
\[ \int_0^{\frac{\pi}{4}} -\sin(x - \frac{\pi}{4}) \, dx = -\int_0^{\frac{\pi}{4}} \sin(x - \frac{\pi}{4}) \, dx \]
Using the substitution \( u = x - \frac{\pi}{4} \), we have \( du = dx \) and the limits change from \( 0 \) to \( -\frac{\pi}{4} \) and from \( \frac{\pi}{4} \) to \( 0 \):
\[ = -\left[-\cos(u)\right]_{0}^{-\frac{\pi}{4}} = -\left[-\cos(-\frac{\pi}{4}) + \cos(0)\right] = -\left[-\frac{1}{\sqrt{2}} + 1\right] = 1 - \frac{1}{\sqrt{2}} \]

### Step 4: Evaluate the second integral
For \( x \) in the interval \( [\frac{\pi}{4}, \frac{\pi}{2}] \), \( \sin(x - \frac{\pi}{4}) \) is positive:
\[ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin(x - \frac{\pi}{4}) \, dx \]
Using the same substitution \( u = x - \frac{\pi}{4} \):
\[ = \int_{0}^{\frac{\pi}{4}} \sin(u) \, du = -\cos(u) \bigg|_{0}^{\frac{\pi}{4}} = -\left[-\frac{1}{\sqrt{2}} + 1\right] = 1 - \frac{1}{\sqrt{2}} \]

### Step 5: Combine the results
Now, we combine both integrals:
\[ I = \left(1 - \frac{1}{\sqrt{2}}\right) + \left(1 - \frac{1}{\sqrt{2}}\right) = 2 - \frac{2}{\sqrt{2}} = 2 - \sqrt{2} \]

### Final Result
Thus, the value of the integral is:
\[ \int_0^{\frac{\pi}{2}} |\sin(x - \frac{\pi}{4})| \, dx = 2 - \sqrt{2} \]