To evaluate the integral \( I = \int_0^1 \frac{\sqrt{1 - x}}{\sqrt{1 + x}} \, dx \), the following steps are executed:Step 1: Integral Rewrite
The integral is initially stated as:\[I = \int_0^1 \frac{\sqrt{1 - x}}{\sqrt{1 + x}} \, dx.\]Multiplying the numerator and denominator by \( \sqrt{1 - x} \) yields:\[I = \int_0^1 \frac{1 - x}{\sqrt{(1 + x)(1 - x)}} \, dx.\]Step 2: Denominator Simplification
Using the difference of squares identity \( (1 + x)(1 - x) = 1 - x^2 \), the integral becomes:\[I = \int_0^1 \frac{1 - x}{\sqrt{1 - x^2}} \, dx.\]Step 3: Integral Decomposition
The integral is divided into two separate integrals:\[I = \int_0^1 \frac{1}{\sqrt{1 - x^2}} \, dx - \int_0^1 \frac{x}{\sqrt{1 - x^2}} \, dx.\]These are designated as \( I_1 \) and \( I_2 \):\[I_1 = \int_0^1 \frac{1}{\sqrt{1 - x^2}} \, dx, \quad I_2 = \int_0^1 \frac{x}{\sqrt{1 - x^2}} \, dx.\]Step 4: \( I_1 \) Evaluation
The evaluation of \( I_1 \) is as follows:\[I_1 = \sin^{-1}(x) \Big|_0^1 = \sin^{-1}(1) - \sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}.\]Step 5: \( I_2 \) Evaluation
For \( I_2 \), a substitution of \( x = \sin(t) \) is performed, leading to \( dx = \cos(t) dt \). The integration limits adjust from \( x = 0, 1 \) to \( t = 0, \frac{\pi}{2} \):\[I_2 = \int_0^{\frac{\pi}{2}} \frac{\sin(t)}{\sqrt{1 - \sin^2(t)}} \cos(t) \, dt = \int_0^{\frac{\pi}{2}} \sin(t) \, dt.\]This integral evaluates to:\[I_2 = -\cos(t) \Big|_0^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right) + \cos(0) = 0 + 1 = 1.\]Step 6: Result Synthesis
The final result is obtained by combining the evaluations of \( I_1 \) and \( I_2 \):\[I = I_1 - I_2 = \frac{\pi}{2} - 1.\]The definitive value of the integral is:\[I = \frac{\pi}{2} - 1.\]