Question

The value of \( \displaystyle \int_{0}^{\pi/2} \frac{\sin x}{\sin x + \cos x} \, dx \) is:

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

Hint:

For definite integrals involving \( \sin x \) and \( \cos x \), try the substitution \( x \rightarrow a-x \). This often converts the integral into a complementary form that simplifies when added.

Answer 1

The Correct Option is C

Concept: For definite integrals of the form \[ I = \int_0^{a} f(x)\,dx \] we can use the property \[ \int_0^{a} f(x)\,dx = \int_0^{a} f(a-x)\,dx \] Adding the two expressions often simplifies symmetric expressions involving \( \sin x \) and \( \cos x \).

Step 1: Let the given integral be \[ I = \int_{0}^{\pi/2} \frac{\sin x}{\sin x + \cos x} \, dx \] Using the property \( x \rightarrow \frac{\pi}{2}-x \): \[ I = \int_{0}^{\pi/2} \frac{\sin\left(\frac{\pi}{2}-x\right)} {\sin\left(\frac{\pi}{2}-x\right)+\cos\left(\frac{\pi}{2}-x\right)}\,dx \] Using identities \[ \sin\left(\frac{\pi}{2}-x\right)=\cos x, \qquad \cos\left(\frac{\pi}{2}-x\right)=\sin x \] Thus, \[ I = \int_{0}^{\pi/2} \frac{\cos x}{\sin x+\cos x} \, dx \]

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

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