Question

The value of the definite integral \(\int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} dx\) is:

• A. \(\pi/2\)
• B. \(\pi/4\)
• C. \(0\)
• D. \(1\)

Hint:

Integrals of the form \(\int_0^{\pi/2} \frac{f(\sin x)}{f(\sin x) + f(\cos x)} dx\) or \(\int_0^{\pi/2} \frac{f(\tan x)}{f(\tan x) + f(\cot x)} dx\) almost always evaluate to \(\frac{\pi}{4}\).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to evaluate a definite integral involving trigonometric functions over the interval \([0, \pi/2]\).

Step 2: Key Formula or Approach:
We use the King's property of definite integrals:
\[ \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \]

Step 3: Detailed Explanation:
Let the given integral be \(I\):
\[ I = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x} \, dx \quad \dots \text{(Equation 1)} \] Apply the property by substituting \(x\) with \((\frac{\pi}{2} - x)\):
\[ I = \int_0^{\pi/2} \frac{\sin(\frac{\pi}{2} - x)}{\sin(\frac{\pi}{2} - x) + \cos(\frac{\pi}{2} - x)} \, dx \] Using the complementary angle identities \(\sin(\frac{\pi}{2} - x) = \cos x\) and \(\cos(\frac{\pi}{2} - x) = \sin x\), the integral transforms into:
\[ I = \int_0^{\pi/2} \frac{\cos x}{\cos x + \sin x} \, dx \quad \dots \text{(Equation 2)} \] Now, add Equation 1 and Equation 2:
\[ I + I = \int_0^{\pi/2} \left( \frac{\sin x}{\sin x + \cos x} + \frac{\cos x}{\sin x + \cos x} \right) \, dx \] \[ 2I = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x} \, dx \] \[ 2I = \int_0^{\pi/2} 1 \, dx \] Evaluating this basic integral gives:
\[ 2I = [x]_0^{\pi/2} = \frac{\pi}{2} - 0 = \frac{\pi}{2} \] Finally, divide by 2 to find \(I\):
\[ I = \frac{\pi}{4} \]

Step 4: Final Answer:
The correct choice is (B).