Find $I = \int_0^1 \frac{\sqrt{1-x}}{\sqrt{x} + \sqrt{1-x}} dx$ (rationalising the denominator)
Step-by-Step Solution
Key Concept: Rationalize the denominator by multiplying by the conjugate, then split the resulting integral into manageable parts using trigonometric and standard substitutions.
Rationalize by multiplying by $\frac{\sqrt{x} - \sqrt{1-x}}{\sqrt{x} - \sqrt{1-x}}$: $I = \int_0^1 \frac{\sqrt{1-x}(\sqrt{x} - \sqrt{1-x})}{x - (1-x)} dx = \int_0^1 \frac{\sqrt{x(1-x)} - (1-x)}{2x-1} dx$. Split into two integrals and evaluate: $I = [\sin^{-1}u]_0^1 - \frac{1}{2}\int_0^1 \frac{2x dx}{\sqrt{1-x^2}} = [\sin^{-1}u]_0^1 - \frac{1}{2}[2\sqrt{1-x^2}]_0^1 = \frac{\pi}{2} - 0 + [\sqrt{1-x^2}]_0^1 = \frac{\pi}{2} + 0 - 1 = \frac{\pi}{2} - 1$.
Correct Answer: $\frac{\pi}{2} - 1$