Problem 182 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Evaluate \(\int_{0}^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x\) (A) \(\frac{\pi}{4}\) (B) \(\frac{\pi}{2}\) (C) zero (D) 1

Step-by-Step Solution

Answer

The value of the integral is \(\frac{\pi}{4}\).

Step 1: Understand the Symmetry

Notice that the function inside the integral, \( f(x) = \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \), seems symmetric with respect to \( x = \frac{\pi}{4} \). This is a clue that we can exploit the property of integrals, \[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \] when \( f(x) + f(a-x) \) gives a constant.

Step 2: Transform the Integral

First, apply the substitution \( x = \frac{\pi}{2} - u \), then \( dx = -du \). Change the limits: when \( x = 0 \), \( u = \frac{\pi}{2} \), and when \( x = \frac{\pi}{2} \), \( u = 0 \). Thus, \[ \int_{0}^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx = \int_{\pi / 2}^{0} \frac{\sqrt{\cos u}}{\sqrt{\cos u} + \sqrt{\sin u}} (-du) \] This simplifies to: \[ \int_{0}^{\pi / 2} \frac{\sqrt{\cos u}}{\sqrt{\sin u} + \sqrt{\cos u}} \, du \]

Step 3: Add the Integral and Its Transformation

Add the original integral and the transformed integral: \[ \int_{0}^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx + \int_{0}^{\pi / 2} \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx \] Notice that \[ \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} = 1 \] Which simplifies to: \[ \int_{0}^{\pi / 2} 1 \, dx = \left[ x \right]_{0}^{\pi / 2} = \frac{\pi}{2} \]

Step 4: Evaluate the Original Integral

Since the sum of the original integral and its transformation is \( \frac{\pi}{2} \), and they are symmetric and equal, each one contributes half. So, \[ 2 \times \text{Integral} = \frac{\pi}{2} \] Therefore, the original integral: \[ \int_{0}^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx = \frac{\pi}{4} \]

Key Concepts

Symmetry in Integrals

Symmetry in integrals is a concept where we can use mathematical properties to simplify complex integrals. An integral is symmetric if we can reflect it across an axis and it remains unchanged. This often indicates that the integral can be split into two equal parts, making calculations easier.
A key formula that helps when dealing with symmetric functions over a definite interval from 0 to a is: \[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \].
In our exercise, the function \( f(x) = \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \) exhibits symmetry around \( x = \frac{\pi}{4} \).

• This symmetry indicates that swapping \( x \) with \( a - x \) leaves the integral unchanged.
• Such functions make it feasible to use symmetry to pair two integrals that add up to a constant.

Recognizing symmetry helps us reduce the complexity and effort needed to compute an integral.

Trigonometric Substitution

Trigonometric substitution is a technique used to simplify integrals involving square roots by using trigonometric identities. It involves substituting a trigonometric function for a variable to simplify the expression.
In this problem, we utilize the substitution \( x = \frac{\pi}{2} - u \) to transform the integral's variable. This changes the limits of integration as well, creating a new integral in terms of \( u \):
\[ \int_{0}^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx = \int_{0}^{\pi / 2} \frac{\sqrt{\cos u}}{\sqrt{\sin u} + \sqrt{\cos u}} \, du \]
Here’s the advantage:

• The substitution often transforms the original trigonometric forms into more manageable ones.
• The flipped limits take advantage of relationships such as \( \sin(\frac{\pi}{2} - u) = \cos(u) \) and \( \cos(\frac{\pi}{2} - u) = \sin(u) \).

Such substitutions streamline the integration process by aligning more closely with standard integral properties.

Integral Properties

Understanding the fundamental properties of integrals can simplify computations significantly. Some key properties are additivity, linearity, and symmetry.
In this exercise, after transforming the integral, we make use of addition of integrals:
\[ \int_{0}^{\pi / 2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx + \int_{0}^{\pi / 2} \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx = \int_{0}^{\pi / 2} 1 \, dx \]
The brilliance here lies in the fact that the individual terms add up to 1:

• \( \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} = 1 \)

This is a clever application of integral properties:

• By adding these two functions, they create a constant integrating to \( \frac{\pi}{2} \) over the interval \([0, \frac{\pi}{2}]\).
• By symmetry, each part contributes equally to the result, splitting the final value by half, yielding \( \frac{\pi}{4} \) as the answer for each component.

Knowing and using these properties allows for strategic manipulation of functions within integrals to derive simpler answers.