`int(sinxcosx)/(3sin^(2)x+5cos^(2)x)dx=`

Options

A. `1/4logabs(3sin^(2)x+5cos^(2)x)+c`
B. `-1/4logabs(3sin^(2)x+5cos^(2)x)+c`
C. `1/4logabs(3sin^(2)x-5cos^(2)x)+c`
D. `-1/4logabs(3sin^(2)x-5cos^(2)x)+c`

Answer

To solve the integral

\[ \int \frac{\sin x \cos x}{3 \sin^2 x + 5 \cos^2 x} \, dx, \]

we can follow these steps:

### Step 1: Simplify the Denominator
We start with the expression in the denominator:

\[ 3 \sin^2 x + 5 \cos^2 x. \]

Using the identity \(\cos^2 x = 1 - \sin^2 x\), we can rewrite the denominator:

\[ 3 \sin^2 x + 5 (1 - \sin^2 x) = 3 \sin^2 x + 5 - 5 \sin^2 x = (3 - 5) \sin^2 x + 5 = -2 \sin^2 x + 5. \]

Thus, the integral becomes:

\[ \int \frac{\sin x \cos x}{5 - 2 \sin^2 x} \, dx. \]

### Step 2: Use Substitution
Let us use the substitution:

\[ t = 5 - 2 \sin^2 x. \]

Now, we need to differentiate \(t\) with respect to \(x\):

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

This implies:

\[ dt = -4 \sin x \cos x \, dx \quad \Rightarrow \quad \sin x \cos x \, dx = -\frac{1}{4} dt. \]

### Step 3: Substitute in the Integral
Substituting \(t\) and \(dt\) into the integral gives:

\[ \int \frac{-\frac{1}{4} dt}{t} = -\frac{1}{4} \int \frac{dt}{t}. \]

### Step 4: Integrate
The integral of \(\frac{1}{t}\) is:

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

Thus, we have:

\[ -\frac{1}{4} \ln |t| + C = -\frac{1}{4} \ln |5 - 2 \sin^2 x| + C. \]

### Step 5: Final Answer
We can express the final answer as:

\[ -\frac{1}{4} \ln |5 - 2 \sin^2 x| + C. \]