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

Answer

To solve the definite 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 integrand using the double angle identity
We know that

\[ \sin 2x = 2 \sin x \cos x. \]

Thus, we can express the integrand 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: Simplify the denominator
We can express the denominator as:

\[ 9 + 32 \sin x \cos x = 9 + 16 \cdot 2 \sin x \cos x = 9 + 16 \sin 2x. \]

### Step 3: Substitute \( t = \sin x - \cos x \)
Let \( t = \sin x - \cos x \). Then, we have:

\[ \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right). \]

### Step 4: Change of variable
We differentiate \( t \):

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

### 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: Substitute into the integral
Now, we can rewrite the integral in terms of \( t \):

\[ I = \int_{-1}^{0} \frac{dt}{25 - 16t^2}. \]

### Step 7: Solve the integral
We can use the formula for the integral:

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

Here, \( a^2 = 25 \) so \( a = 5 \):

\[ I = \int_{-1}^{0} \frac{dt}{25 - 16t^2} = \frac{1}{10} \left[ \ln \left| \frac{5 + 4t}{5 - 4t} \right| \right]_{-1}^{0}. \]

### Step 8: Evaluate the limits
Calculating the limits:

At \( t = 0 \):

\[ \frac{5 + 4(0)}{5 - 4(0)} = \frac{5}{5} = 1 \implies \ln(1) = 0. \]

At \( t = -1 \):

\[ \frac{5 + 4(-1)}{5 - 4(-1)} = \frac{5 - 4}{5 + 4} = \frac{1}{9} \implies \ln\left(\frac{1}{9}\right) = -\ln(9). \]

### Final Calculation
Thus, we have:

\[ I = \frac{1}{10} \left[ 0 - (-\ln(9)) \right] = \frac{1}{10} \ln(9) = \frac{1}{10} \cdot 2 \ln(3) = \frac{1}{5} \ln(3). \]

### Conclusion
The final result is:

\[ I = \frac{1}{5} \ln(3). \]