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

Answer

To evaluate the integral

\[ \int_{0}^{1} \frac{-1}{\sqrt{1-x^2}} \, dx, \]

we can follow these steps:

### Step 1: Identify the integral
We have the integral

\[ \int_{0}^{1} \frac{-1}{\sqrt{1-x^2}} \, dx. \]

### Step 2: Use the known integral formula
We know from integral calculus that

\[ \int \frac{1}{\sqrt{1-x^2}} \, dx = \sin^{-1}(x) + C. \]

Thus,

\[ \int \frac{-1}{\sqrt{1-x^2}} \, dx = -\sin^{-1}(x) + C. \]

### Step 3: Evaluate the definite integral
Now we will evaluate the definite integral from 0 to 1:

\[ \int_{0}^{1} \frac{-1}{\sqrt{1-x^2}} \, dx = \left[-\sin^{-1}(x)\right]_{0}^{1}. \]

### Step 4: Calculate the values at the limits
Now we substitute the limits into the expression:

1. At \(x = 1\):

\[ -\sin^{-1}(1) = -\frac{\pi}{2}. \]

2. At \(x = 0\):

\[ -\sin^{-1}(0) = 0. \]

### Step 5: Combine the results
Now, we combine the results from the limits:

\[ \left[-\sin^{-1}(x)\right]_{0}^{1} = -\frac{\pi}{2} - 0 = -\frac{\pi}{2}. \]

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

\[ -\frac{\pi}{2}. \]

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