Question

\textbf \( \int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx \) is equal to:}

• A. \( 0 \)
• B. \( -\pi \)
• C. \( \frac{3\pi}{2} \)
• D. \( \frac{\pi}{2} \)
• E. \( \frac{\pi}{4} \)

Hint:

For integrals of the form \( \int_{0}^{\pi/2} \frac{f(\sin x)}{f(\sin x) + f(\cos x)} \, dx \), the result is always half of the upper limit, which is \( \pi/4 \). This applies to powers, roots, or even logarithmic functions of sin and cos.

Answer 1

Answer: (E)

Concept: This problem is solved using the property of definite integrals: \( \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx \). This is often called "King's Property." By applying this, we can create a second integral that, when added to the first, simplifies the integrand significantly.

Step 1: Apply the integral property.
Let \( I = \int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx \quad \cdots (1) \)
Using the property \( x \rightarrow \frac{\pi}{2} - x \): \[ I = \int_{0}^{\pi/2} \frac{\sqrt{\sin(\pi/2 - x)}}{\sqrt{\sin(\pi/2 - x)} + \sqrt{\cos(\pi/2 - x)}} \, dx \] Since \( \sin(\pi/2 - x) = \cos x \) and \( \cos(\pi/2 - x) = \sin x \): \[ I = \int_{0}^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} \, dx \quad \cdots (2) \]

Step 2: Add the two integrals.
Adding (1) and (2): \[ 2I = \int_{0}^{\pi/2} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx \] \[ 2I = \int_{0}^{\pi/2} 1 \, dx = [x]_{0}^{\pi/2} = \frac{\pi}{2} \] \[ I = \frac{\pi}{4} \]