Question

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

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

Hint:

For definite integrals of the form \( \int_0^{a} f(x)\,dx \), always check the symmetry property \( f(x) + f(a-x) \). If their sum simplifies to a constant, the integral becomes very easy to evaluate.

Answer 1

The Correct Option is C

Concept: For definite integrals over symmetric limits, an important property is: \[ \int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx \] This property is particularly useful when the function contains expressions like \( \sin x \) and \( \cos x \), since under the transformation \( x \to \frac{\pi}{2}-x \): \[ \sin\left(\frac{\pi}{2}-x\right)=\cos x, \qquad \cos\left(\frac{\pi}{2}-x\right)=\sin x \] Using this symmetry often simplifies integrals significantly.

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

Step 2: Use the property \( \int_a^b f(x)dx=\int_a^b f(a+b-x)dx \). Here \(a=0\) and \(b=\frac{\pi}{2}\). Replace \(x\) with \( \frac{\pi}{2}-x \): \[ I=\int_{0}^{\pi/2} \frac{\sqrt{\sin\left(\frac{\pi}{2}-x\right)}} {\sqrt{\sin\left(\frac{\pi}{2}-x\right)}+\sqrt{\cos\left(\frac{\pi}{2}-x\right)}}\,dx \] Using trigonometric identities: \[ \sin\left(\frac{\pi}{2}-x\right)=\cos x, \qquad \cos\left(\frac{\pi}{2}-x\right)=\sin x \] Thus, \[ I=\int_{0}^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx \]

Step 3: Add the two expressions for \(I\). \[ 2I= \int_{0}^{\pi/2} \left( \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} \right)dx \] Since the denominators are the same: \[ \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} =1 \] Therefore, \[ 2I=\int_{0}^{\pi/2} 1\,dx \]

Step 4: Evaluate the remaining integral. \[ 2I=\left[x\right]_{0}^{\pi/2} =\frac{\pi}{2} \] \[ I=\frac{\pi}{4} \]