Question

Evaluate \( \displaystyle \int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx \).

• A. \( \frac{\pi}{2} \)
• B. \( \frac{\pi}{6} \)
• C. \( \frac{\pi}{4} \)
• D. \( \frac{\pi}{3} \)

Hint:

For definite integrals of the form \( \int_0^{a} f(x)\,dx \), always check the symmetry property \( f(x) + f(a-x) \). If their sum simplifies to a constant, the integral becomes very easy to evaluate.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem uses the property of definite integrals where $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx$. This is particularly useful for trigonometric functions with limits involving $\pi/2$.
Step 2: Key Formula or Approach:
Let the given integral be $I$. We apply the property: \[ I = \int_{0}^{\pi/2} \frac{\sqrt{\sin(\frac{\pi}{2}-x)}}{\sqrt{\sin(\frac{\pi}{2}-x)} + \sqrt{\cos(\frac{\pi}{2}-x)}} \, dx \]
Step 3: Detailed Explanation:
Since $\sin(\frac{\pi}{2}-x) = \cos x$ and $\cos(\frac{\pi}{2}-x) = \sin x$, we have: \[ I = \int_{0}^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} \, dx \quad \dots (1) \] The original integral is: \[ I = \int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx \quad \dots (2) \] Adding (1) and (2): \[ 2I = \int_{0}^{\pi/2} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx \] \[ 2I = \int_{0}^{\pi/2} 1 \, dx = [x]_{0}^{\pi/2} = \frac{\pi}{2} \] \[ I = \frac{\pi}{4} \]
Step 4: Final Answer:
The value of the integral is $\frac{\pi}{4}$.