`int(sin x +8 cosx)/(4 sin x +6cosx)dx=`

Options

A. `x+(1)/(2) log | 4 sin x +6 cos x|+C`
B. `2x +log |2 sin x+3 cos x|+C`
C. `x+2 log |2 sin x+3 cos x|+C`
D. `(1)/(2) log |4 sin x+6 cos x | +C`

Explanation

To solve the integral \(\int \frac{\sin x + 8 \cos x}{4 \sin x + 6 \cos x} \, dx\), we will use the method of partial fractions. Here is a step-by-step solution:

### Step 1: Set up the equation
We can express the numerator \(\sin x + 8 \cos x\) as a linear combination of the derivative of the denominator \(4 \sin x + 6 \cos x\) and the denominator itself. Thus, we can write:
\[ \sin x + 8 \cos x = a(4 \cos x - 6 \sin x) + b(4 \sin x + 6 \cos x) \]
where \(a\) and \(b\) are constants to be determined.

### Step 2: Expand the equation
Expanding the right-hand side:
\[ \sin x + 8 \cos x = (4b + 6a) \cos x + (-6a + 4b) \sin x \]

### Step 3: Equate coefficients
Now we equate the coefficients of \(\sin x\) and \(\cos x\) from both sides:
1. For \(\sin x\): \(1 = -6a + 4b\) (Equation 1)
2. For \(\cos x\): \(8 = 4a + 6b\) (Equation 2)

### Step 4: Solve the system of equations
Now we solve the system of equations formed by Equation 1 and Equation 2.

From Equation 1:
\[ 6a - 4b = -1 \quad \text{(1)} \]
From Equation 2:
\[ 4a + 6b = 8 \quad \text{(2)} \]

Multiplying Equation (1) by 3:
\[ 18a - 12b = -3 \quad \text{(3)} \]

Multiplying Equation (2) by 2:
\[ 8a + 12b = 16 \quad \text{(4)} \]

Now, add Equations (3) and (4):
\[ (18a + 8a) + (-12b + 12b) = -3 + 16 \]
\[ 26a = 13 \implies a = \frac{1}{2} \]

Substituting \(a = \frac{1}{2}\) into Equation (1):
\[ 1 = -6\left(\frac{1}{2}\right) + 4b \]
\[ 1 = -3 + 4b \implies 4b = 4 \implies b = 1 \]

### Step 5: Substitute back into the equation
Now substituting \(a\) and \(b\) back into our expression:
\[ \sin x + 8 \cos x = \frac{1}{2}(4 \cos x - 6 \sin x) + 1(4 \sin x + 6 \cos x) \]
This simplifies to:
\[ \sin x + 8 \cos x = 2 \cos x - 3 \sin x + 4 \sin x + 6 \cos x = 8 \cos x + \sin x \]
This confirms our values for \(a\) and \(b\).

### Step 6: Rewrite the integral
Now we can rewrite the integral:
\[ \int \frac{\sin x + 8 \cos x}{4 \sin x + 6 \cos x} \, dx = \int \left( \frac{1}{2} \frac{4 \cos x - 6 \sin x}{4 \sin x + 6 \cos x} + \frac{4 \sin x + 6 \cos x}{4 \sin x + 6 \cos x} \right) \, dx \]
This simplifies to:
\[ \frac{1}{2} \int \frac{4 \cos x - 6 \sin x}{4 \sin x + 6 \cos x} \, dx + \int dx \]

### Step 7: Change of variables
Let \(t = 4 \sin x + 6 \cos x\). Then, differentiating:
\[ dt = (4 \cos x - 6 \sin x) \, dx \]
Thus:
\[ \int \frac{4 \cos x - 6 \sin x}{4 \sin x + 6 \cos x} \, dx = \int \frac{1}{t} \, dt = \ln |t| + C \]

### Step 8: Final integration
Now substituting back:
\[ \frac{1}{2} \ln |4 \sin x + 6 \cos x| + x + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{\sin x + 8 \cos x}{4 \sin x + 6 \cos x} \, dx = \frac{1}{2} \ln |4 \sin x + 6 \cos x| + x + C \]