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

Concept: For definite integrals, an important property is \[ \int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a-x)\,dx \] This symmetry property is often useful when the integrand contains both \(\sin x\) and \(\cos x\).

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

Step 2: Add the two integrals. \[ 2I=\int_{0}^{\pi/2} \frac{\sin x+\cos x}{\sin x+\cos x}\,dx \] \[ 2I=\int_{0}^{\pi/2} 1\,dx \] \[ 2I=\frac{\pi}{2} \]

Step 3: Solve for \(I\). \[ I=\frac{\pi}{4} \]