`int(sin2x)/(1+sin^(2)x)dx=`

Options

A. `log(sin2x)+c`
B. `log(1+sin^(2)x)+c`
C. `1/2log(1+sin^(2)x)+c`
D. `tan^(-1)(sinx)+c`

Answer

To solve the integral \( \int \frac{\sin(2x)}{1 + \sin^2(x)} \, dx \), we can use the substitution method. Here’s a step-by-step solution:

### Step 1: Identify the substitution
We notice that the derivative of \( 1 + \sin^2(x) \) involves \( \sin(2x) \). Therefore, we can let:
\[ t = 1 + \sin^2(x) \]

### Step 2: Differentiate the substitution
Next, we differentiate \( t \) with respect to \( x \):
\[ \frac{dt}{dx} = 0 + 2\sin(x)\cos(x) = \sin(2x) \]
Thus, we can express \( dt \) as:
\[ dt = \sin(2x) \, dx \]

### Step 3: Rewrite the integral
Now we can rewrite the integral in terms of \( t \):
\[ \int \frac{\sin(2x)}{1 + \sin^2(x)} \, dx = \int \frac{\sin(2x)}{t} \cdot \frac{dt}{\sin(2x)} = \int \frac{1}{t} \, dt \]

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

### Step 5: Substitute back
Now we substitute back \( t = 1 + \sin^2(x) \):
\[ \ln |1 + \sin^2(x)| + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{\sin(2x)}{1 + \sin^2(x)} \, dx = \ln(1 + \sin^2(x)) + C \]

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