`int(sinx+cosx)/(3sinx+4cosx)dx=`

Options

A. `(1)/(25)(11x+2log|3sinx+4cosx|)+c`
B. `(1)/(25)(11x-2log|3sinx+4cosx|)+c`
C. `(1)/(25)(11x+log|3sinx+4cosx|)+c`
D. `(1)/(25)(11x-log|3sinx+4cosx|)+c`

Answer

To solve the integral \(\int \frac{\sin x + \cos x}{3 \sin x + 4 \cos x} \, dx\), we can follow these steps:

### Step 1: Set up the integral
Let
\[ I = \int \frac{\sin x + \cos x}{3 \sin x + 4 \cos x} \, dx. \]

### Step 2: Identify functions
Let:
- \( f(x) = 3 \sin x + 4 \cos x \)
- \( g(x) = \sin x + \cos x \)

We will express \( g(x) \) in terms of \( f(x) \) and its derivative.

### Step 3: Find the derivative of \( f(x) \)
Calculating the derivative of \( f(x) \):
\[ f'(x) = 3 \cos x - 4 \sin x. \]

### Step 4: Set up the equation
We want to express \( g(x) \) as:
\[ g(x) = a f(x) + b f'(x). \]
Substituting \( f(x) \) and \( f'(x) \):
\[ \sin x + \cos x = a(3 \sin x + 4 \cos x) + b(3 \cos x - 4 \sin x). \]

### Step 5: Expand and collect terms
Expanding the right-hand side:
\[ \sin x + \cos x = (3a - 4b) \sin x + (4a + 3b) \cos x. \]

### Step 6: Set up a system of equations
By comparing coefficients, we have:
1. \( 3a - 4b = 1 \) (coefficient of \(\sin x\))
2. \( 4a + 3b = 1 \) (coefficient of \(\cos x\))

### Step 7: Solve the system of equations
From the first equation:
\[ 3a - 4b = 1 \quad \text{(1)} \]
From the second equation:
\[ 4a + 3b = 1 \quad \text{(2)} \]

Multiply equation (1) by 3:
\[ 9a - 12b = 3 \quad \text{(3)} \]

Multiply equation (2) by 4:
\[ 16a + 12b = 4 \quad \text{(4)} \]

Adding equations (3) and (4):
\[ 25a = 7 \implies a = \frac{7}{25}. \]

Substituting \( a \) back into equation (1):
\[ 3\left(\frac{7}{25}\right) - 4b = 1 \implies \frac{21}{25} - 4b = 1 \implies -4b = 1 - \frac{21}{25} = -\frac{4}{25} \implies b = \frac{1}{25}. \]

### Step 8: Substitute back into the integral
Now we have \( a = \frac{7}{25} \) and \( b = \frac{1}{25} \). Therefore, we can rewrite the integral as:
\[ I = \int \left( \frac{7}{25}(3 \sin x + 4 \cos x) + \frac{1}{25}(3 \cos x - 4 \sin x) \right) \frac{dx}{3 \sin x + 4 \cos x}. \]

### Step 9: Separate the integral
This can be separated into two integrals:
\[ I = \frac{7}{25} \int 1 \, dx + \frac{1}{25} \int \frac{3 \cos x - 4 \sin x}{3 \sin x + 4 \cos x} \, dx. \]

### Step 10: Solve the integrals
The first integral is straightforward:
\[ \frac{7}{25} x. \]

For the second integral, let \( u = 3 \sin x + 4 \cos x \), then \( du = (3 \cos x - 4 \sin x) dx \). Thus,
\[ \int \frac{du}{u} = \ln |u| + C = \ln |3 \sin x + 4 \cos x| + C. \]

### Final Result
Combining both parts, we get:
\[ I = \frac{7}{25} x - \frac{1}{25} \ln |3 \sin x + 4 \cos x| + C. \]