Consider the integrals `I=int(sinx)/(3cos x+4sinx)dx and J=int(cosx)/(3cos x+4sinx)dx`. Then, the integral `4J-3J` is equal to (where, c is the constant of integration)

Options

A. `ln|2 cos x-4 sinx|+c`
B. `ln|3 cos x+4 sin x|+c`
C. `ln|3 sin x+4 cos x|+c`
D. `ln|3 sin x-4 cos x|+c`

Answer

To solve the problem, we need to evaluate the integrals \( I = \int \frac{\sin x}{3 \cos x + 4 \sin x} \, dx \) and \( J = \int \frac{\cos x}{3 \cos x + 4 \sin x} \, dx \), and then find \( 4J - 3I \).

### Step 1: Write down the integrals
We have:
\[ I = \int \frac{\sin x}{3 \cos x + 4 \sin x} \, dx \]
\[ J = \int \frac{\cos x}{3 \cos x + 4 \sin x} \, dx \]

### Step 2: Calculate \( 4J - 3I \)
We can express \( 4J - 3I \) as follows:
\[ 4J - 3I = 4 \int \frac{\cos x}{3 \cos x + 4 \sin x} \, dx - 3 \int \frac{\sin x}{3 \cos x + 4 \sin x} \, dx \]
Combining the integrals gives:
\[ 4J - 3I = \int \left( \frac{4 \cos x - 3 \sin x}{3 \cos x + 4 \sin x} \right) \, dx \]

### Step 3: Simplify the expression
Now, we can simplify the integral:
\[ 4J - 3I = \int \frac{4 \cos x - 3 \sin x}{3 \cos x + 4 \sin x} \, dx \]

### Step 4: Substitute for easier integration
Let \( t = 3 \cos x + 4 \sin x \). Then, we differentiate \( t \):
\[ dt = (-3 \sin x + 4 \cos x) \, dx \]
This means:
\[ dx = \frac{dt}{-3 \sin x + 4 \cos x} \]

### Step 5: Rewrite the integral in terms of \( t \)
We can express the integral as:
\[ \int \frac{4 \cos x - 3 \sin x}{t} \cdot \frac{dt}{-3 \sin x + 4 \cos x} \]
Notice that \( 4 \cos x - 3 \sin x \) can be rewritten to match the derivative of \( t \):
\[ 4 \cos x - 3 \sin x = -\frac{4}{3} dt \]

### Step 6: Integrate
Now we have:
\[ 4J - 3I = \int \frac{dt}{t} \]
The integral of \( \frac{1}{t} \) is:
\[ \ln |t| + C \]

### Step 7: Substitute back for \( t \)
Substituting back for \( t \):
\[ 4J - 3I = \ln |3 \cos x + 4 \sin x| + C \]

### Final Result
Thus, the final answer is:
\[ 4J - 3I = \ln |3 \cos x + 4 \sin x| + C \]