Consider the integrals `I=int(sinx)/(3cos x+4sinx)dx and J=int(cosx)/(3cos x+4sinx)dx`. Then, the integral `4J-3I` 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 expression \( 4J - 3I \) where:

\[ I = \int \frac{\sin x}{3 \cos x + 4 \sin x} \, dx \]
\[ J = \int \frac{\cos x}{3 \cos x + 4 \sin x} \, dx \]

### Step 1: Write the expression for \( 4J - 3I \)

We can express \( 4J - 3I \) as follows:

\[ 4J - 3I = 4 \int \frac{\cos x}{3 \cos x + 4 \sin x} \, dx - 3 \int \frac{\sin x}{3 \cos x + 4 \sin x} \, dx \]

### Step 2: Combine the integrals

Since both integrals have the same denominator, we can combine them:

\[ 4J - 3I = \int \left( \frac{4 \cos x - 3 \sin x}{3 \cos x + 4 \sin x} \right) \, dx \]

### Step 3: Substitution

Let \( t = 3 \cos x + 4 \sin x \). Then, we differentiate \( t \):

\[ dt = (-3 \sin x + 4 \cos x) \, dx \]

Rearranging gives us:

\[ dx = \frac{dt}{-3 \sin x + 4 \cos x} \]

### Step 4: Rewrite the integral

Now, we need to express \( 4 \cos x - 3 \sin x \) in terms of \( t \):

From our substitution, we can express \( 4 \cos x - 3 \sin x \) as:

\[ 4 \cos x - 3 \sin x = \frac{4 \cos x - 3 \sin x}{-3 \sin x + 4 \cos x} \cdot (-3 \sin x + 4 \cos x) \]

Thus, we can rewrite our integral:

\[ 4J - 3I = \int \frac{4 \cos x - 3 \sin x}{3 \cos x + 4 \sin x} \cdot \frac{dt}{-3 \sin x + 4 \cos x} \]

### Step 5: Simplify the integral

Notice that:

\[ \frac{4 \cos x - 3 \sin x}{-3 \sin x + 4 \cos x} = -1 \]

Thus, we can simplify our integral:

\[ 4J - 3I = -\int \frac{dt}{t} \]

### Step 6: Evaluate the integral

The integral of \( \frac{1}{t} \) is:

\[ -\int \frac{dt}{t} = -\ln |t| + C \]

Substituting back for \( t \):

\[ 4J - 3I = -\ln |3 \cos x + 4 \sin x| + C \]

### Final Answer

Thus, the value of \( 4J - 3I \) is:

\[ 4J - 3I = -\ln |3 \cos x + 4 \sin x| + C \]