Question

The value of the integral \[ \int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx \] is:

• A. \( \frac{\pi}{2} \)
• B. \( -\frac{\pi}{2} \)
• C. \( \frac{\pi}{4} \)
• D. None of these

Hint:

For integrals with symmetry, use substitution and combine the integrals to simplify the evaluation.

Answer 1

The Correct Option is C

Step 1: We can use symmetry to evaluate the integral. Let’s define: \[ I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx. \] Step 2: We can use the substitution \( x = \frac{\pi}{2} - t \). Then, \( dx = -dt \) and the limits of integration change as follows: when \( x = 0 \), \( t = \frac{\pi}{2} \), and when \( x = \frac{\pi}{2} \), \( t = 0 \). Substituting into the integral: \[ I = \int_{\frac{\pi}{2}}^0 \frac{\sqrt{\sin \left( \frac{\pi}{2} - t \right)}}{\sqrt{\sin \left( \frac{\pi}{2} - t \right)} + \sqrt{\cos \left( \frac{\pi}{2} - t \right)}} \, (-dt). \] Since \( \sin \left( \frac{\pi}{2} - t \right) = \cos t \) and \( \cos \left( \frac{\pi}{2} - t \right) = \sin t \), the integral becomes: \[ I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\cos t}}{\sqrt{\cos t} + \sqrt{\sin t}} \, dt. \] Step 3: Adding the original and transformed integrals: \[ 2I = \int_0^{\frac{\pi}{2}} \left( \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} \right) \, dx. \] The sum of the two fractions is 1, so: \[ 2I = \int_0^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2}. \] Thus: \[ I = \frac{\pi}{4}. \]