Question

The value of the integral \( \int_0^1 \frac{\sqrt{1 - x}}{\sqrt{1 + x}} \, dx \) is:

• A. \( \frac{\pi}{2} + 1 \)
• B. \( \frac{\pi}{2} - 1 \)
• C. \( 1 \)
• D. \( -1 \)

Hint:

Use trigonometric substitution to simplify integrals with square roots involving \( 1 - x \) and \( 1 + x \).

Answer 1

The Correct Option is B

To solve the integral \( I = \int_0^1 \frac{\sqrt{1 - x}}{\sqrt{1 + x}} \, dx \), we follow these steps: Step 1: Rewrite the Integral
We start with the integral: \[ I = \int_0^1 \frac{\sqrt{1 - x}}{\sqrt{1 + x}} \, dx. \] To simplify this, we can multiply and divide by \( \sqrt{1 - x} \): \[ I = \int_0^1 \frac{1 - x}{\sqrt{(1 + x)(1 - x)}} \, dx. \] Step 2: Simplify the Denominator
The denominator can be simplified using the identity \( (1 + x)(1 - x) = 1 - x^2 \): \[ I = \int_0^1 \frac{1 - x}{\sqrt{1 - x^2}} \, dx. \] Step 3: Split the Integral
We can split the integral into two parts: \[ I = \int_0^1 \frac{1}{\sqrt{1 - x^2}} \, dx - \int_0^1 \frac{x}{\sqrt{1 - x^2}} \, dx. \] Let’s denote these two integrals 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: Evaluate \( I_1 \)
The integral \( I_1 \) is known to be: \[ I_1 = \sin^{-1}(x) \Big|_0^1 = \sin^{-1}(1) - \sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}. \] Step 5: Evaluate \( I_2 \)
For \( I_2 \), we use the substitution \( x = \sin(t) \), hence \( dx = \cos(t) dt \). The limits change from \( x = 0 \) to \( x = 1 \), which corresponds to \( t = 0 \) to \( t = \frac{\pi}{2} \): \[ I_2 = \int_0^{\frac{\pi}{2}} \frac{\sin(t)}{\sqrt{1 - \sin^2(t)}} \cos(t) \, dt = \int_0^{\frac{\pi}{2}} \sin(t) \, dt. \] This evaluates to: \[ I_2 = -\cos(t) \Big|_0^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right) + \cos(0) = 0 + 1 = 1. \] Step 6: Combine Results
Now we can combine the results of \( I_1 \) and \( I_2 \): \[ I = I_1 - I_2 = \frac{\pi}{2} - 1. \] Thus, the value of the integral is: \[ I = \frac{\pi}{2} - 1. \]