Question

The value of \( \int_{0}^{\pi/2} \frac{dx}{1 + \sqrt{\tan x}} \) will be

• A. 0
• B. \( \frac{\pi}{2} \)
• C. \( \frac{\pi}{4} \)
• D. \( \frac{\pi}{8} \)

Hint:

Recognize this standard form. Any integral of the type \( \int_{0}^{\pi/2} \frac{f(x)}{f(x) + f(\pi/2 - x)} \,dx \) is equal to \( \frac{\pi}{4} \). In this case, if we let \( g(x) = \sqrt{\tan x} \), the denominator is \( 1 + g(x) \). This might not seem to fit directly, but after converting to sine and cosine, the form \( \int_{0}^{\pi/2} \frac{h(\cos x)}{h(\cos x) + h(\sin x)} \,dx \) appears, which also evaluates to \( \frac{\pi}{4} \).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This is a standard problem in definite integrals that can be solved efficiently using a property of definite integrals, often known as the "King's Rule".
Step 2: Key Formula or Approach:
The property states that for a continuous function \(f(x)\) on \( [0, a] \): \[ \int_{0}^{a} f(x) \,dx = \int_{0}^{a} f(a-x) \,dx \] Step 3: Detailed Calculation:
Let the given integral be \( I \). \[ I = \int_{0}^{\pi/2} \frac{dx}{1 + \sqrt{\tan x}} \] First, express \( \tan x \) in terms of \( \sin x \) and \( \cos x \): \[ I = \int_{0}^{\pi/2} \frac{dx}{1 + \frac{\sqrt{\sin x}}{\sqrt{\cos x}}} = \int_{0}^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} \,dx .(1) \] Now, apply the property \( \int_{0}^{a} f(x) \,dx = \int_{0}^{a} f(a-x) \,dx \) with \( a = \pi/2 \). \[ I = \int_{0}^{\pi/2} \frac{\sqrt{\cos(\frac{\pi}{2}-x)}}{\sqrt{\cos(\frac{\pi}{2}-x)} + \sqrt{\sin(\frac{\pi}{2}-x)}} \,dx \] Using the trigonometric identities \( \cos(\frac{\pi}{2}-x) = \sin x \) and \( \sin(\frac{\pi}{2}-x) = \cos x \), we get: \[ I = \int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \,dx .(2) \] Now, add equation (1) and equation (2): \[ I + I = \int_{0}^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} \,dx + \int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \,dx \] \[ 2I = \int_{0}^{\pi/2} \frac{\sqrt{\cos x} + \sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}} \,dx \] \[ 2I = \int_{0}^{\pi/2} 1 \,dx \] \[ 2I = [x]_{0}^{\pi/2} \] \[ 2I = \frac{\pi}{2} - 0 = \frac{\pi}{2} \] \[ I = \frac{\pi}{4} \] Step 4: Final Answer:
The value of the integral is \( \frac{\pi}{4} \).