Question

Evaluate the integral \[ \int_{0}^{\pi/2} \frac{\sin x}{\sin x + \cos x}\,dx. \]

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

Hint:

When a definite integral contains \(\sin x\) and \(\cos x\) together, try the property \[ \int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a-x)\,dx. \] Adding the two integrals often simplifies the expression dramatically.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem asks us to evaluate a definite integral involving trigonometric functions with limits from \(0\) to \(\pi/2\).
This type of integral is classic and is best solved using properties of definite integrals rather than direct integration.
Step 2: Key Formula or Approach:
The most useful property for limits between \(0\) and \(a\) is known as King's Rule:
\[ \int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a-x)\,dx \] Using this property typically transforms the numerator to match a part of the denominator, allowing us to add the original and transformed integrals.
Step 3: Detailed Solution:
Let the given integral be denoted as \(I\):
\[ I = \int_{0}^{\pi/2} \frac{\sin x}{\sin x+\cos x}\,dx \quad \dots \text{(Equation 1)} \] Applying the definite integral property \(\int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a-x)\,dx\):
\[ I = \int_{0}^{\pi/2} \frac{\sin(\frac{\pi}{2}-x)}{\sin(\frac{\pi}{2}-x)+\cos(\frac{\pi}{2}-x)}\,dx \] Since \(\sin(\frac{\pi}{2}-x) = \cos x\) and \(\cos(\frac{\pi}{2}-x) = \sin x\), the integral becomes:
\[ I = \int_{0}^{\pi/2} \frac{\cos x}{\cos x+\sin x}\,dx \quad \dots \text{(Equation 2)} \] Now, add Equation 1 and Equation 2:
\[ I + I = \int_{0}^{\pi/2} \frac{\sin x}{\sin x+\cos x}\,dx + \int_{0}^{\pi/2} \frac{\cos x}{\sin x+\cos x}\,dx \] \[ 2I = \int_{0}^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x}\,dx \] The integrand simplifies to \(1\):
\[ 2I = \int_{0}^{\pi/2} 1\,dx \] \[ 2I = \left[ x \right]_{0}^{\pi/2} \] \[ 2I = \frac{\pi}{2} - 0 \] \[ 2I = \frac{\pi}{2} \] Solving for \(I\):
\[ I = \frac{\pi}{4} \] Step 4: Final Answer:
The evaluated integral is \(\frac{\pi}{4}\).