Evaluate: `int_(0)^(1) (1)/(sqrt(3+2x-x^(2))dx`

Answer

To evaluate the integral

\[ I = \int_{0}^{1} \frac{1}{\sqrt{3 + 2x - x^2}} \, dx, \]

we will manipulate the expression under the square root to make it easier to integrate.

### Step 1: Rewrite the expression under the square root

We start with the expression \(3 + 2x - x^2\). We can rearrange it as follows:

\[ 3 + 2x - x^2 = - (x^2 - 2x - 3). \]

Next, we factor the quadratic:

\[ x^2 - 2x - 3 = (x - 3)(x + 1). \]

Thus, we have:

\[ 3 + 2x - x^2 = -((x - 3)(x + 1)). \]

### Step 2: Complete the square

To make the expression a perfect square, we can rewrite it as:

\[ 3 + 2x - x^2 = 4 - (x - 1)^2. \]

This is because:

\[ 4 - (x - 1)^2 = 4 - (x^2 - 2x + 1) = 3 + 2x - x^2. \]

### Step 3: Substitute into the integral

Now we substitute this back into the integral:

\[ I = \int_{0}^{1} \frac{1}{\sqrt{4 - (x - 1)^2}} \, dx. \]

### Step 4: Use trigonometric substitution

We can use the substitution \(x - 1 = 2 \sin \theta\), which implies \(dx = 2 \cos \theta \, d\theta\). The limits change as follows:

- When \(x = 0\), \(2 \sin \theta = -1\) which gives \(\theta = -\frac{\pi}{6}\).
- When \(x = 1\), \(2 \sin \theta = 0\) which gives \(\theta = 0\).

Now substituting into the integral:

\[ I = \int_{-\frac{\pi}{6}}^{0} \frac{2 \cos \theta}{\sqrt{4 - (2 \sin \theta)^2}} \, d\theta = \int_{-\frac{\pi}{6}}^{0} \frac{2 \cos \theta}{\sqrt{4(1 - \sin^2 \theta)}} \, d\theta = \int_{-\frac{\pi}{6}}^{0} \frac{2 \cos \theta}{\sqrt{4 \cos^2 \theta}} \, d\theta. \]

This simplifies to:

\[ I = \int_{-\frac{\pi}{6}}^{0} \frac{2 \cos \theta}{2 \cos \theta} \, d\theta = \int_{-\frac{\pi}{6}}^{0} 1 \, d\theta. \]

### Step 5: Evaluate the integral

Now we can evaluate the integral:

\[ I = \left[ \theta \right]_{-\frac{\pi}{6}}^{0} = 0 - \left(-\frac{\pi}{6}\right) = \frac{\pi}{6}. \]

### Final Answer

Thus, the value of the integral is:

\[ \boxed{\frac{\pi}{6}}. \]