`int_(0)^(1) x sqrt((1-x^(2))/(1+x^(2)))dx`

Answer

To solve the integral \( I = \int_{0}^{1} x \sqrt{\frac{1-x^2}{1+x^2}} \, dx \), we can follow these steps:

### Step 1: Substitution
Let \( x^2 = t \). Then, we have:
\[ dx = \frac{1}{2\sqrt{t}} \, dt \]
When \( x = 0 \), \( t = 0 \) and when \( x = 1 \), \( t = 1 \). Thus, the integral becomes:
\[ I = \int_{0}^{1} \sqrt{t} \sqrt{\frac{1-t}{1+t}} \cdot \frac{1}{2\sqrt{t}} \, dt = \frac{1}{2} \int_{0}^{1} \sqrt{\frac{1-t}{1+t}} \, dt \]

### Step 2: Rationalizing the Integral
Now, we can rewrite the integral:
\[ I = \frac{1}{2} \int_{0}^{1} \frac{\sqrt{1-t}}{\sqrt{1+t}} \, dt \]
To simplify this, we can rationalize the denominator:
\[ \sqrt{1+t} = \sqrt{(1-t)(1+t)} \Rightarrow \sqrt{1-t} \cdot \sqrt{1+t} \]
Thus, we have:
\[ I = \frac{1}{2} \int_{0}^{1} \frac{(1-t)}{\sqrt{1-t^2}} \, dt \]

### Step 3: Splitting the Integral
We can separate the integral:
\[ I = \frac{1}{2} \left( \int_{0}^{1} \frac{1}{\sqrt{1-t^2}} \, dt - \int_{0}^{1} \frac{t}{\sqrt{1-t^2}} \, dt \right) \]

### Step 4: Evaluating the First Integral
The first integral \( \int_{0}^{1} \frac{1}{\sqrt{1-t^2}} \, dt \) is known:
\[ \int_{0}^{1} \frac{1}{\sqrt{1-t^2}} \, dt = \frac{\pi}{2} \]

### Step 5: Evaluating the Second Integral
For the second integral, we can use the substitution \( t = \sin(\theta) \), which gives \( dt = \cos(\theta) \, d\theta \):
\[ \int_{0}^{1} \frac{t}{\sqrt{1-t^2}} \, dt = \int_{0}^{\frac{\pi}{2}} \sin(\theta) \, d\theta = 1 \]

### Step 6: Combining Results
Now we can combine the results:
\[ I = \frac{1}{2} \left( \frac{\pi}{2} - 1 \right) = \frac{\pi}{4} - \frac{1}{2} \]

### Final Result
Thus, the value of the integral is:
\[ I = \frac{\pi}{4} - \frac{1}{2} \]