Question

\( \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx \) is equal to:

• A. \( \frac{\pi}{4} \)
• B. \( \frac{\pi}{6} \)
• C. \( \frac{\pi}{12} \)
• D. \( \frac{\pi}{2} \)

Hint:

For any integral of the form \( \int_a^b \frac{f(x)}{f(x) + f(a+b-x)} dx \), the shortcut answer is always \( \frac{b - a}{2} \).

Answer 1

The Correct Option is C

Step 1: Key Formula or Approach:
Use the definite integral property: \( \int_a^b f(x) dx = \int_a^b f(a + b - x) dx \).
Step 2: Detailed Explanation:
Let \( I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx \) ... (i)
Here \( a = \pi/6 \) and \( b = \pi/3 \). The sum \( a + b = \pi/6 + \pi/3 = \pi/2 \).
Applying the property, replace \( x \) with \( \pi/2 - x \):
\[ I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos(\pi/2 - x)}}{\sqrt{\sin(\pi/2 - x)} + \sqrt{\cos(\pi/2 - x)}} dx \] \[ I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx \) ... (ii)
Add equations (i) and (ii):
\[ 2I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x} + \sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx \] \[ 2I = \int_{\pi/6}^{\pi/3} 1 \, dx = [x]_{\pi/6}^{\pi/3} \] \[ 2I = \frac{\pi}{3} - \frac{\pi}{6} = \frac{2\pi - \pi}{6} = \frac{\pi}{6} \] \[ I = \frac{\pi}{12} \] Step 3: Final Answer:
The value of the integral is \( \frac{\pi}{12} \).