Compute the integrals
`I = int_(0)^(pi//4) (sin x + cos x)/( 3 + sin 2 x) dx`

Answer

To compute the integral

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{3 + \sin 2x} \, dx, \]

we can follow these steps:

### Step 1: Simplify the Integral
First, we can rewrite \(\sin 2x\) using the double angle identity:
\[ \sin 2x = 2 \sin x \cos x. \]
Thus, the integral becomes:
\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{3 + 2 \sin x \cos x} \, dx. \]

### Step 2: Substitution
Let \(t = \sin x + \cos x\). Then, we differentiate \(t\) to find \(dt\):
\[ dt = \cos x \, dx - \sin x \, dx = (\cos x - \sin x) \, dx. \]
This means:
\[ dx = \frac{dt}{\cos x - \sin x}. \]

### Step 3: Change the Limits
Now we need to change the limits of integration. When \(x = 0\):
\[ t = \sin(0) + \cos(0) = 1. \]
When \(x = \frac{\pi}{4}\):
\[ t = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}. \]
So the new limits are from \(1\) to \(\sqrt{2}\).

### Step 4: Substitute in the Integral
Now we substitute \(t\) into the integral:
\[ I = \int_{1}^{\sqrt{2}} \frac{t}{3 + 2(\frac{t^2 - 1}{2})} \cdot \frac{dt}{\cos x - \sin x}. \]
We can express \(\cos x - \sin x\) in terms of \(t\):
\[ \cos x - \sin x = \sqrt{2} \cos\left(x + \frac{\pi}{4}\right). \]
However, for simplicity, we will keep the integral in terms of \(t\).

### Step 5: Simplify the Denominator
The denominator simplifies as follows:
\[ 3 + 2 \sin x \cos x = 3 + (t^2 - 1) = t^2 + 2. \]
Thus, the integral becomes:
\[ I = \int_{1}^{\sqrt{2}} \frac{t}{t^2 + 2} \, dt. \]

### Step 6: Integration
Now we can integrate:
\[ I = \int \frac{t}{t^2 + 2} \, dt. \]
Using the substitution \(u = t^2 + 2\), we have \(du = 2t \, dt\) or \(dt = \frac{du}{2t}\). Thus:
\[ I = \frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln |u| + C = \frac{1}{2} \ln |t^2 + 2| + C. \]

### Step 7: Evaluate the Integral
Now we evaluate from \(1\) to \(\sqrt{2}\):
\[ I = \frac{1}{2} \left[ \ln(\sqrt{2}^2 + 2) - \ln(1^2 + 2) \right] = \frac{1}{2} \left[ \ln(2 + 2) - \ln(1 + 2) \right] = \frac{1}{2} \left[ \ln(4) - \ln(3) \right]. \]
This simplifies to:
\[ I = \frac{1}{2} \ln\left(\frac{4}{3}\right). \]

### Final Answer
Thus, the value of the integral is:
\[ I = \frac{1}{2} \ln\left(\frac{4}{3}\right). \]