Question

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

• A. \( \pi \)
• B. \( \pi/2 \)
• C. \( \pi/4 \)
• D. \( 0 \)

Hint:

For integrals involving \( \sin x \) and \( \cos x \) over \( [0,\pi/2] \), try using the symmetry property \( x \rightarrow \frac{\pi}{2}-x \). Adding the two forms of the integral often simplifies the expression dramatically.

Answer 1

The Correct Option is C

Concept: A useful symmetry property of definite integrals is: \[ \int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx \] This property helps simplify integrals containing complementary trigonometric functions like \( \sin x \) and \( \cos x \), especially over intervals such as \( [0,\pi/2] \). Another useful trick is adding the integral with its transformed form to simplify the expression.

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

Step 2: Using the property \( \int_0^a f(x)dx = \int_0^a f(a-x)dx \). Let \( x \rightarrow \frac{\pi}{2} - x \). Since \[ \sin\left(\frac{\pi}{2}-x\right) = \cos x \] and \[ \cos\left(\frac{\pi}{2}-x\right) = \sin x \] we get \[ I = \int_{0}^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} \, dx \]

Step 3: Adding the two expressions of \(I\). \[ I = \int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx \] \[ I = \int_{0}^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx \] Adding, \[ 2I = \int_{0}^{\pi/2} \left( \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} \right) dx \] \[ 2I = \int_{0}^{\pi/2} 1 \, dx \] \[ 2I = \left[x\right]_{0}^{\pi/2} = \frac{\pi}{2} \] \[ I = \frac{\pi}{4} \]