If `int f(x) sin x cos x dx = 1/(2(b^2 - a^2)) log f(x) + c` then `f(x) =`

Options

A. `1/(a^2 sin^2 x + b^2 cos^2 x)`
B. `1/(a sin x + b sin x)`
C. `1/(a^2 cos^2 x + b^2 sin^2 x)`
D. none

Answer

To solve the equation given in the question, we start with the integral equation:

\[ \int f(x) \sin x \cos x \, dx = \frac{1}{2(b^2 - a^2)} \log f(x) + c \]

### Step 1: Differentiate both sides with respect to \(x\)

We differentiate the left-hand side using the product rule and the right-hand side using the chain rule:

\[ \frac{d}{dx} \left( \int f(x) \sin x \cos x \, dx \right) = f(x) \sin x \cos x \]

For the right-hand side, we have:

\[ \frac{d}{dx} \left( \frac{1}{2(b^2 - a^2)} \log f(x) + c \right) = \frac{1}{2(b^2 - a^2)} \cdot \frac{1}{f(x)} \cdot f'(x) \]

### Step 2: Set the derivatives equal to each other

Equating both sides gives us:

\[ f(x) \sin x \cos x = \frac{1}{2(b^2 - a^2)} \cdot \frac{1}{f(x)} \cdot f'(x) \]

### Step 3: Multiply both sides by \(f(x)\)

To eliminate the fraction, we multiply both sides by \(f(x)\):

\[ f(x)^2 \sin x \cos x = \frac{1}{2(b^2 - a^2)} f'(x) \]

### Step 4: Rearrange the equation

Rearranging gives us:

\[ f'(x) = 2(b^2 - a^2) f(x)^2 \sin x \cos x \]

### Step 5: Separate variables

We can separate the variables:

\[ \frac{f'(x)}{f(x)^2} = 2(b^2 - a^2) \sin x \cos x \]

### Step 6: Integrate both sides

Integrating both sides:

\[ \int \frac{f'(x)}{f(x)^2} \, dx = \int 2(b^2 - a^2) \sin x \cos x \, dx \]

The left side integrates to:

\[ -\frac{1}{f(x)} \]

The right side can be simplified using the identity \(\sin x \cos x = \frac{1}{2} \sin(2x)\):

\[ \int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) \]

Thus, we have:

\[ -\frac{1}{f(x)} = (b^2 - a^2)(-\frac{1}{2} \cos(2x)) + C \]

### Step 7: Solve for \(f(x)\)

Rearranging gives:

\[ \frac{1}{f(x)} = \frac{(b^2 - a^2)}{2} \cos(2x) - C \]

Taking the reciprocal:

\[ f(x) = \frac{1}{\frac{(b^2 - a^2)}{2} \cos(2x) - C} \]

### Final Result

Thus, the function \(f(x)\) is:

\[ f(x) = \frac{2}{(b^2 - a^2) \cos(2x) - 2C} \]