Evaluate : `int_(0)^((pi)/(2))(sin 2x)/(1+sin^(2) x)dx`

Answer

To evaluate the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin 2x}{1 + \sin^2 x} \, dx, \]

we can follow these steps:

### Step 1: Simplify the Integral
We start with the integral as given:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin 2x}{1 + \sin^2 x} \, dx. \]

### Step 2: Use a Substitution
We can use the substitution \( t = 1 + \sin^2 x \). To find \( dt \), we differentiate:

\[ dt = 2\sin x \cos x \, dx = \sin 2x \, dx. \]

This means that \( \sin 2x \, dx = dt \).

### Step 3: Change the Limits of Integration
Next, we need to change the limits of integration based on our substitution.

- When \( x = 0 \):
\[ t = 1 + \sin^2(0) = 1 + 0 = 1. \]

- When \( x = \frac{\pi}{2} \):
\[ t = 1 + \sin^2\left(\frac{\pi}{2}\right) = 1 + 1 = 2. \]

### Step 4: Rewrite the Integral
Now we can rewrite the integral in terms of \( t \):

\[ I = \int_{1}^{2} \frac{dt}{t}. \]

### Step 5: Evaluate the Integral
The integral of \( \frac{1}{t} \) is \( \ln t \):

\[ I = \left[ \ln t \right]_{1}^{2} = \ln 2 - \ln 1. \]

Since \( \ln 1 = 0 \), we have:

\[ I = \ln 2. \]

### Final Answer
Thus, the value of the integral is

\[ \boxed{\ln 2}. \]

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