`int (sin 2x)/(1+sin^2x)dx=`

Options

A. `log|sin 2x|+c`
B. `log|1+sin^2x|+c`
C. `(1)/(2)log|1+sin^2 x|+c`
D. ` tan^(-1) (sin x)+c`

Answer

To solve the integral \( I = \int \frac{\sin 2x}{1 + \sin^2 x} \, dx \), we can follow these steps:

### Step 1: Rewrite the integrand
We know that \( \sin 2x = 2 \sin x \cos x \). Therefore, we can rewrite the integral as:
\[ I = \int \frac{2 \sin x \cos x}{1 + \sin^2 x} \, dx \]

### Step 2: Use substitution
Let \( t = 1 + \sin^2 x \). Then, we need to find \( dt \):
\[ dt = 2 \sin x \cos x \, dx \]
This means:
\[ dx = \frac{dt}{2 \sin x \cos x} \]

### Step 3: Substitute into the integral
Now substituting \( t \) into the integral:
\[ I = \int \frac{2 \sin x \cos x}{t} \cdot \frac{dt}{2 \sin x \cos x} \]
The \( 2 \sin x \cos x \) cancels out:
\[ I = \int \frac{1}{t} \, dt \]

### Step 4: Integrate
The integral \( \int \frac{1}{t} \, dt \) is:
\[ I = \ln |t| + C \]

### Step 5: Substitute back for \( t \)
Now, substituting back for \( t \):
\[ I = \ln |1 + \sin^2 x| + C \]

### Final Answer
Thus, the final answer is:
\[ I = \ln |1 + \sin^2 x| + C \]