Question

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

• A. \(\frac{\pi}{2}\)
• B. \(\frac{\pi}{8}\)
• C. \(\frac{\pi}{4}\)
• D. \(1\)

Hint:

For integrals of the form \(\int_0^{\pi/2} \frac{f(\sin x,\cos x)}{g(\sin x,\cos x)}dx\), try the substitution \(x \to \frac{\pi}{2}-x\) and add both expressions.

Answer 1

The Correct Option is C

Topic - Definite Integrals (Properties):
This is a classic integral problem solvable using the properties of definite integrals, often referred to as "King's Rule".
Step 1: Understanding the Question:
The objective is to evaluate the definite integral of a trigonometric fraction over the interval \([0, \pi/2]\).
Step 2: Key Formula or Approach:
Use the property: \[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx \] Step 3: Detailed Solution:
1. Let \(I = \int_{0}^{\pi/2} \frac{\sin x}{\sin x + \cos x} \, dx\) --- (Eq. 1).
2. Apply the property \(x \to \pi/2 - x\):
\[ I = \int_{0}^{\pi/2} \frac{\sin(\pi/2 - x)}{\sin(\pi/2 - x) + \cos(\pi/2 - x)} \, dx \] \[ I = \int_{0}^{\pi/2} \frac{\cos x}{\cos x + \sin x} \, dx \] --- (Eq. 2).
3. Add (Eq. 1) and (Eq. 2):
\[ 2I = \int_{0}^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x} \, dx \] \[ 2I = \int_{0}^{\pi/2} 1 \, dx \] 4. Integrate:
\[ 2I = [x]_{0}^{\pi/2} = \frac{\pi}{2} \] \[ I = \frac{\pi}{4} \] Step 4: Final Answer:
The value of the integral is \(\frac{\pi}{4}\).