Question

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

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

Hint:

When a definite integral contains symmetric expressions like \(\sin x\) and \(\cos x\), applying the property \[ \int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx \] often simplifies the integral significantly.

Answer 1

The Correct Option is C

Concept: A useful property of definite integrals is: \[ \int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx \] This symmetry property helps simplify integrals when expressions contain complementary trigonometric terms such as \(\sin x\) and \(\cos x\).

Step 1: Let the given integral be \(I\). \[ I = \int_{0}^{\pi/2} \frac{\sin x}{\sin x + \cos x}\,dx \] Using the property \[ \int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx \] we write \[ I = \int_{0}^{\pi/2} \frac{\sin\left(\frac{\pi}{2}-x\right)}{\sin\left(\frac{\pi}{2}-x\right)+\cos\left(\frac{\pi}{2}-x\right)}\,dx \]

Step 2: Use trigonometric identities. \[ \sin\left(\frac{\pi}{2}-x\right) = \cos x \] \[ \cos\left(\frac{\pi}{2}-x\right) = \sin x \] Thus, \[ I = \int_{0}^{\pi/2} \frac{\cos x}{\sin x + \cos x}\,dx \]

Step 3: Add the two expressions of \(I\). \[ I = \int_{0}^{\pi/2} \frac{\sin x}{\sin x+\cos x}\,dx \] \[ I = \int_{0}^{\pi/2} \frac{\cos x}{\sin x+\cos x}\,dx \] Adding, \[ 2I = \int_{0}^{\pi/2} \left( \frac{\sin x}{\sin x+\cos x} + \frac{\cos x}{\sin x+\cos x} \right) dx \] \[ 2I = \int_{0}^{\pi/2} 1\,dx \] \[ 2I = \left[x\right]_{0}^{\pi/2} = \frac{\pi}{2} \]

Step 4: Solve for \(I\). \[ I = \frac{\pi}{4} \]