`int_(0)^(pi//4)(sin x + cos x)/(9+16 sin 2 x )`

Answer

To solve the integral

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{9 + 16 \sin 2x} \, dx \]

we will follow these steps:

### Step 1: Rewrite the integral
We start by rewriting the integral in a more manageable form. We know that \( \sin 2x = 2 \sin x \cos x \). Thus, we can express the integral as:

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{9 + 16(2 \sin x \cos x)} \, dx = \int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{9 + 32 \sin x \cos x} \, dx \]

### Step 2: Factor out constants
Next, we can factor out a constant from the denominator:

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{16 \left( \frac{9}{16} + 2 \sin x \cos x \right)} \, dx = \frac{1}{16} \int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{\frac{9}{16} + 2 \sin x \cos x} \, dx \]

### Step 3: Use a trigonometric identity
Using the identity \( \sin^2 x + \cos^2 x = 1 \), we can rewrite \( 2 \sin x \cos x \) as \( \sin 2x \):

\[ I = \frac{1}{16} \int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{\frac{9}{16} + \sin 2x} \, dx \]

### Step 4: Substitute \( t = \sin x - \cos x \)
Let \( t = \sin x - \cos x \). Then, we have:

\[ dt = (\cos x + \sin x) \, dx \]

### Step 5: Change the limits of integration
When \( x = 0 \), \( t = \sin(0) - \cos(0) = -1 \). When \( x = \frac{\pi}{4} \), \( t = \sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right) = 0 \).

### Step 6: Rewrite the integral in terms of \( t \)
Now we can rewrite the integral in terms of \( t \):

\[ I = \frac{1}{16} \int_{-1}^{0} \frac{dt}{\frac{9}{16} + \sin 2x} \]

### Step 7: Solve the integral
Using the formula for the integral of the form \( \int \frac{1}{a^2 - x^2} \, dx \):

\[ \int \frac{1}{a^2 - x^2} \, dx = \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right| + C \]

Here, \( a = \frac{5}{4} \), we compute:

\[ I = \frac{1}{16} \cdot \frac{1}{2 \cdot \frac{5}{4}} \left[ \ln \left| \frac{\frac{5}{4} + t}{\frac{5}{4} - t} \right| \right]_{-1}^{0} \]

### Step 8: Evaluate the limits
Calculate the value of the integral at the limits:

\[ I = \frac{1}{40} \left( \ln \left| \frac{\frac{5}{4} + 0}{\frac{5}{4} - 0} \right| - \ln \left| \frac{\frac{5}{4} - (-1)}{\frac{5}{4} + 1} \right| \right) \]

### Step 9: Simplify the expression
This simplifies to:

\[ I = \frac{1}{40} \left( 0 - \ln \left| \frac{\frac{9}{4}}{\frac{9}{4}} \right| \right) \]

### Final Result
Thus, the value of the integral is:

\[ I = \frac{1}{20} \ln 3 \]