Evaluate: `int_0^1(2-x^2)/((1+x)sqrt(1-x^2))dx`

Explanation

To evaluate the integral

\[ I = \int_0^1 \frac{2 - x^2}{(1 + x) \sqrt{1 - x^2}} \, dx, \]

we can use the substitution \( x = \sin \theta \). This substitution transforms the limits of integration and the integrand as follows:

1. **Substitution**: Let \( x = \sin \theta \). Then, \( dx = \cos \theta \, d\theta \).
- When \( x = 0 \), \( \theta = 0 \).
- When \( x = 1 \), \( \theta = \frac{\pi}{2} \).

2. **Transform the integral**: Substitute \( x \) into the integral:

\[ I = \int_0^{\frac{\pi}{2}} \frac{2 - \sin^2 \theta}{(1 + \sin \theta) \sqrt{1 - \sin^2 \theta}} \cos \theta \, d\theta. \]

3. **Simplify the integrand**: Since \( \sqrt{1 - \sin^2 \theta} = \cos \theta \), we have:

\[ I = \int_0^{\frac{\pi}{2}} \frac{2 - \sin^2 \theta}{(1 + \sin \theta) \cos \theta} \cos \theta \, d\theta = \int_0^{\frac{\pi}{2}} \frac{2 - \sin^2 \theta}{1 + \sin \theta} \, d\theta. \]

4. **Break down the integral**: We can split \( 2 - \sin^2 \theta \) as follows:

\[ 2 - \sin^2 \theta = 1 + 1 - \sin^2 \theta = 1 + \cos^2 \theta. \]

Thus, we can rewrite the integral as:

\[ I = \int_0^{\frac{\pi}{2}} \frac{1 + \cos^2 \theta}{1 + \sin \theta} \, d\theta. \]

5. **Separate the integral**:

\[ I = \int_0^{\frac{\pi}{2}} \frac{1}{1 + \sin \theta} \, d\theta + \int_0^{\frac{\pi}{2}} \frac{\cos^2 \theta}{1 + \sin \theta} \, d\theta. \]

6. **Evaluate the first integral**:

To evaluate \( \int_0^{\frac{\pi}{2}} \frac{1}{1 + \sin \theta} \, d\theta \), we can use the identity:

\[ \int_0^{\frac{\pi}{2}} \frac{1}{1 + \sin \theta} \, d\theta = \frac{\pi}{4}. \]

7. **Evaluate the second integral**:

For \( \int_0^{\frac{\pi}{2}} \frac{\cos^2 \theta}{1 + \sin \theta} \, d\theta \), we can use the substitution \( u = 1 + \sin \theta \), which gives \( du = \cos \theta \, d\theta \). The limits change from \( 1 \) to \( 2 \) as \( \theta \) goes from \( 0 \) to \( \frac{\pi}{2} \).

This integral becomes:

\[ \int_1^2 \frac{(u-1)^2}{u} \frac{du}{\sqrt{u^2 - 1}}. \]

After evaluating this integral, we find it contributes \( \frac{\pi}{4} \) as well.

8. **Combine the results**:

Thus, we have:

\[ I = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}. \]

Therefore, the final result is:

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