Evaluate :
`int (sin (2 tan^(-1)x))/(1+x^(2))dx`

Answer

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

### Step 1: Substitution
Let \( t = \tan^{-1} x \). Then, the derivative of \( t \) with respect to \( x \) is:
\[ \frac{dt}{dx} = \frac{1}{1 + x^2} \]
Thus, we can express \( dx \) in terms of \( dt \):
\[ dx = (1 + x^2) \, dt \]

### Step 2: Rewrite the Integral
Substituting \( t = \tan^{-1} x \) into the integral, we have:
\[ \sin(2 \tan^{-1} x) = \sin(2t) \]
So the integral becomes:
\[ \int \frac{\sin(2t)}{1 + x^2} \, dx = \int \sin(2t) \, dt \]

### Step 3: Integrate
The integral of \( \sin(2t) \) is:
\[ \int \sin(2t) \, dt = -\frac{1}{2} \cos(2t) + C \]

### Step 4: Back Substitute
Now, we substitute back \( t = \tan^{-1} x \):
\[ -\frac{1}{2} \cos(2 \tan^{-1} x) + C \]

### Final Answer
Thus, the final answer is:
\[ -\frac{1}{2} \cos(2 \tan^{-1} x) + C \]

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