Evaluate the definite integrals `int_(0)^(pi//4)(sinx+cosx)/(25-16(sinx-cosx)^(2)) dx`

Explanation

To evaluate the definite integral

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{25 - 16(\sin x - \cos x)^2} \, dx, \]

we will perform a substitution and simplify the integral step by step.

### Step 1: Substitution
Let \( t = \sin x - \cos x \). Then, we differentiate to find \( dt \):
\[ dt = (\cos x + \sin x) \, dx. \]
Thus, we have:
\[ dx = \frac{dt}{\cos x + \sin x}. \]

### Step 2: Change the limits of integration
Next, we need to change the limits of integration according to our substitution:
- When \( x = 0 \):
\[ t = \sin(0) - \cos(0) = 0 - 1 = -1. \]
- When \( x = \frac{\pi}{4} \):
\[ t = \sin\left(\frac{\pi}{4}\right) - \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 0. \]

Thus, the new limits for \( t \) will be from \( -1 \) to \( 0 \).

### Step 3: Rewrite the integral
Now, we can rewrite the integral in terms of \( t \):
\[ I = \int_{-1}^{0} \frac{\sin x + \cos x}{25 - 16t^2} \cdot \frac{dt}{\sin x + \cos x}. \]
The \( \sin x + \cos x \) terms cancel out:
\[ I = \int_{-1}^{0} \frac{1}{25 - 16t^2} \, dt. \]

### Step 4: Simplify the integral
We can rewrite the denominator:
\[ 25 - 16t^2 = 5^2 - (4t)^2. \]
Now we can use 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, \( a = 5 \) and \( x = 4t \).

### Step 5: Apply the formula
Thus,
\[ I = \frac{1}{2 \cdot 5} \int_{-1}^{0} \frac{1}{5^2 - (4t)^2} \, dt = \frac{1}{10} \int_{-1}^{0} \frac{1}{25 - 16t^2} \, dt. \]
Using the formula:
\[ = \frac{1}{10} \cdot \frac{1}{10} \left[ \ln \left| \frac{5 + 4t}{5 - 4t} \right| \right]_{-1}^{0}. \]

### Step 6: Evaluate the limits
Now we evaluate the limits:
- For \( t = 0 \):
\[ \frac{5 + 4(0)}{5 - 4(0)} = \frac{5}{5} = 1 \implies \ln(1) = 0. \]
- For \( t = -1 \):
\[ \frac{5 + 4(-1)}{5 - 4(-1)} = \frac{5 - 4}{5 + 4} = \frac{1}{9} \implies \ln\left(\frac{1}{9}\right) = -\ln(9). \]

Thus,
\[ I = \frac{1}{100} \left[ 0 - (-\ln(9)) \right] = \frac{\ln(9)}{100}. \]

### Step 7: Final simplification
Using properties of logarithms:
\[ \ln(9) = \ln(3^2) = 2\ln(3). \]
Thus,
\[ I = \frac{2\ln(3)}{100} = \frac{\ln(3)}{50}. \]

### Final Answer
The value of the definite integral is:
\[ \boxed{\frac{\ln(3)}{50}}. \]