<p>The value of \(I = \int_0^{\pi/2} \dfrac{(\sin x + \cos x)^2}{\sqrt{1 + \sin 2x}}\, dx\) is</p>
Step-by-Step Solution
Key Concept: Recognize that (sin x + cos x)² = 1 + sin 2x, so the integrand simplifies to √(1 + sin 2x). Then rewrite 1 + sin 2x = sin²x + cos²x + 2sin x cos x = (sin x + cos x)² to get √((sin x + cos x)²) = |sin x + cos x|.
<p><strong>Step 1:</strong> Expand the numerator: (sin x + cos x)² = sin²x + cos²x + 2sin x cos x = 1 + sin 2x</p><p><strong>Step 2:</strong> The integrand becomes: $\frac{1 + \sin 2x}{\sqrt{1 + \sin 2x}} = \frac{(\sin x + \cos x)^2}{|\sin x + \cos x|}$</p><p><strong>Step 3:</strong> Since sin x + cos x > 0 on [0, π/2], we have: $\frac{(\sin x + \cos x)^2}{\sin x + \cos x} = \sin x + \cos x$</p><p><strong>Step 4:</strong> Integrate: $I = \int_0^{\pi/2} (\sin x + \cos x)\, dx = [-\cos x + \sin x]_0^{\pi/2}$</p><p><strong>Step 5:</strong> Evaluate: $I = [(-\cos(\pi/2) + \sin(\pi/2)) - (-\cos 0 + \sin 0)] = (0 + 1) - (-1 + 0) = 1 + 1 = 2$</p><p>∴ Answer: C (which equals 2)</p>
Correct Answer: C