Question

Find the value of the integral \( \int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx \).

• A. \( \pi \)
• B. \( \pi/2 \)
• C. \( \pi/4 \)
• D. \( 0 \)

Hint:

For integrals involving \( \sin x \) and \( \cos x \) over \( [0,\pi/2] \), try using the symmetry property \( x \rightarrow \frac{\pi}{2}-x \). Adding the two forms of the integral often simplifies the expression dramatically.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The objective is to evaluate a definite integral involving trigonometric functions in the denominator and numerator over the interval from \( 0 \) to \( \pi/2 \).
The structure of the integrand suggests the use of symmetry properties to simplify the calculation.
Step 2: Key Formula or Approach:
The primary tool used here is the Reflection Property (also known as King's Property) of definite integrals:
\[ \int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx \]
For the interval \( [0, \pi/2] \), the substitution is \( x \to \pi/2 - x \).
Step 3: Detailed Explanation:
Let the given integral be \( I \):
\[ I = \int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx \quad \dots (1) \]
Applying the property \( \int_0^a f(x) \, dx = \int_0^a f(a-x) \, dx \), we replace \( x \) with \( \pi/2 - x \):
\[ I = \int_{0}^{\pi/2} \frac{\sqrt{\sin(\pi/2 - x)}}{\sqrt{\sin(\pi/2 - x)} + \sqrt{\cos(\pi/2 - x)}} \, dx \]
Using the trigonometric identities \( \sin(\pi/2 - x) = \cos x \) and \( \cos(\pi/2 - x) = \sin x \), we get:
\[ I = \int_{0}^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} \, dx \quad \dots (2) \]
Now, add equation (1) and equation (2):
\[ 2I = \int_{0}^{\pi/2} \left( \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \right) \, dx \]
\[ 2I = \int_{0}^{\pi/2} 1 \, dx \]
\[ 2I = [x]_0^{\pi/2} = \frac{\pi}{2} - 0 \]
\[ I = \frac{\pi}{4} \]
Step 4: Final Answer:
The value of the integral is \( \pi/4 \).