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

Answer

To solve the integral \( I = \int \frac{\sin 2x}{1 + \sin^2 x} \, dx \), we will use substitution and integration techniques. Let's go through the steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{\sin 2x}{1 + \sin^2 x} \, dx \]
Recall that \( \sin 2x = 2 \sin x \cos x \). 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 \( u = \sin x \). Then, the derivative \( du = \cos x \, dx \) or \( dx = \frac{du}{\cos x} \). We also note that \( \cos x = \sqrt{1 - u^2} \) (since \( \sin^2 x + \cos^2 x = 1 \)).

Substituting these into the integral gives:
\[ I = \int \frac{2u \sqrt{1 - u^2}}{1 + u^2} \cdot \frac{du}{\sqrt{1 - u^2}} \]
This simplifies to:
\[ I = 2 \int \frac{u}{1 + u^2} \, du \]

### Step 3: Integrate
Now we can integrate:
\[ I = 2 \int \frac{u}{1 + u^2} \, du \]
Using the substitution \( v = 1 + u^2 \), we have \( dv = 2u \, du \) or \( du = \frac{dv}{2u} \). Thus:
\[ I = 2 \cdot \frac{1}{2} \int \frac{1}{v} \, dv = \int \frac{1}{v} \, dv = \ln |v| + C = \ln |1 + u^2| + C \]

### Step 4: Back Substitute
Now we substitute back \( u = \sin x \):
\[ I = \ln |1 + \sin^2 x| + C \]

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