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

Answer

To solve the integral

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

we will follow these steps:

### Step 1: Identify the Integral
The integral we need to evaluate is

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

This is a standard integral that can be solved using the formula:

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

where \( a = 1 \) in our case.

### Step 2: Apply the Formula
Using the formula, we have:

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

### Step 3: Evaluate the Definite Integral
Since we are dealing with a definite integral from 0 to 1, we will evaluate:

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

### Step 4: Substitute the Limits
Now we will substitute the limits into the evaluated integral:

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

### Step 5: Calculate the Values
We know that:

- \(\sin^{-1}(1) = \frac{\pi}{2}\)
- \(\sin^{-1}(0) = 0\)

Thus, we have:

\[ \frac{\pi}{2} - 0 = \frac{\pi}{2}. \]

### Final Answer
Therefore, the value of the integral is:

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

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