Prove that `int_(-pi//4)^(pi//4) log(sinx + cosx)\ dx = -pi/4\ log2.`

Explanation

To prove that

\[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \log(\sin x + \cos x) \, dx = -\frac{\pi}{4} \log 2, \]

we will proceed step by step.

### Step 1: Define the Integral
Let

\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \log(\sin x + \cos x) \, dx. \]

### Step 2: Rewrite the Argument of the Logarithm
We can multiply and divide the argument of the logarithm by \(\sqrt{2}\):

\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \log\left(\frac{\sqrt{2}}{\sqrt{2}}(\sin x + \cos x)\right) \, dx = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \log\left(\sqrt{2} \left(\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x\right)\right) \, dx. \]

### Step 3: Simplify the Logarithm
Using the property of logarithms \(\log(ab) = \log a + \log b\):

\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \log(\sqrt{2}) \, dx + \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \log\left(\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x\right) \, dx. \]

### Step 4: Evaluate the First Integral
The first integral can be calculated as:

\[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \log(\sqrt{2}) \, dx = \log(\sqrt{2}) \cdot \left(\frac{\pi}{4} - \left(-\frac{\pi}{4}\right)\right) = \log(\sqrt{2}) \cdot \frac{\pi}{2} = \frac{\pi}{2} \cdot \frac{1}{2} \log(2) = \frac{\pi}{4} \log(2). \]

### Step 5: Change of Variable for the Second Integral
Now we focus on the second integral. Let \(t = x + \frac{\pi}{4}\), then \(dx = dt\). When \(x = -\frac{\pi}{4}\), \(t = 0\) and when \(x = \frac{\pi}{4}\), \(t = \frac{\pi}{2}\). Thus:

\[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \log\left(\frac{1}{\sqrt{2}} \sin x + \frac{1}{\sqrt{2}} \cos x\right) \, dx = \int_{0}^{\frac{\pi}{2}} \log\left(\frac{1}{\sqrt{2}} \sin(t - \frac{\pi}{4})\right) \, dt. \]

Using the angle subtraction formula:

\[ \sin\left(t - \frac{\pi}{4}\right) = \sin t \cos\left(\frac{\pi}{4}\right) - \cos t \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}(\sin t - \cos t). \]

Thus, the integral becomes:

\[ \int_{0}^{\frac{\pi}{2}} \log\left(\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}(\sin t - \cos t)\right) \, dt = \int_{0}^{\frac{\pi}{2}} \log\left(\frac{1}{2}(\sin t - \cos t)\right) \, dt. \]

### Step 6: Use Known Result
We know that:

\[ \int_{0}^{\frac{\pi}{2}} \log(\sin t) \, dt = -\frac{\pi}{2} \log(2). \]

Thus, we can conclude:

\[ I = \frac{\pi}{4} \log(2) - \frac{\pi}{2} \log(2) = -\frac{\pi}{4} \log(2). \]

### Final Result
Therefore, we have proved that:

\[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \log(\sin x + \cos x) \, dx = -\frac{\pi}{4} \log 2. \]