`int(sin2xdx)/(1+sin^(2)x)`

Options

A. `tan^(-1)(sinx)+c`
B. `tan^(-1)(cosx)+c`
C. `log(1+sin^(2)x)+c`
D. `log(x+sin^(2)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 \(\sin 2x\)
Using the double angle identity, we can express \(\sin 2x\) as:
\[ \sin 2x = 2 \sin x \cos x \]
Thus, we can rewrite the integral as:
\[ I = \int \frac{2 \sin x \cos x}{1 + \sin^2 x} \, dx \]

### Step 2: Use substitution
Let us make the substitution:
\[ t = 1 + \sin^2 x \]
Now, we need to find \(dt\). The derivative of \(t\) with respect to \(x\) is:
\[ dt = 2 \sin x \cos x \, dx \]
This implies:
\[ dx = \frac{dt}{2 \sin x \cos x} \]

### Step 3: Substitute in the integral
Substituting \(t\) and \(dx\) into the integral, we have:
\[ I = \int \frac{2 \sin x \cos x}{t} \cdot \frac{dt}{2 \sin x \cos x} \]
The \(2 \sin x \cos x\) terms cancel out, leaving us with:
\[ I = \int \frac{1}{t} \, dt \]

### Step 4: Integrate
The integral of \(\frac{1}{t}\) is:
\[ I = \log |t| + C \]
Substituting back for \(t\):
\[ I = \log |1 + \sin^2 x| + C \]

### Final Answer
Thus, the final result of the integral is:
\[ I = \log(1 + \sin^2 x) + C \]