Problem 225 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Statement-I: The value of the integral \(\int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\tan x}}\) is equal to \(\frac{\pi}{6}\). Statement-II: \(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x .\) [2013] (A) Statement-I is True; Statement-II is true; Statement-II is not a correct explanation for Statement-I (B) Statement-I is True; Statement-II is False. (C) Statement-I is False; Statement-II is True (D) Statement-I is True; Statement-II is True; Statement-II is a correct explanation for Statement-I

Step-by-Step Solution

Answer

D: Both statements are true, and II explains I.

Step 1: Analyzing the Integral

Consider the integral: \( I = \int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{\tan x}} \). It might simplify using a substitution or a property of definite integrals.

Step 2: Applying the Symmetry Property

Statement-II introduces a property: \( \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx \). This property is sometimes useful in simplifying integrals and can help if applied correctly to this problem.

Step 3: Symmetry Check for Statement-I

Notice that \( \pi/6 + \pi/3 = \pi/2 \). If we apply the property to the given integral: \[ \int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{\tan x}} = \int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{\tan(\pi/2 - x)}} \]Since \( \tan(\frac{\pi}{2} - x) = \cot x \), the function transforms into \( \int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{\cot x}} \).

Step 4: Evaluate the Transformed Integral

Combining the results from the symmetry:\[ I = \int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{\tan x}} + \int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{\cot x}} \]Since \( \frac{1}{1+\sqrt{\tan x}} + \frac{1}{1+\frac{1}{\sqrt{\tan x}}} = 1 \), integrating gives:\\( I + I = \frac{\pi}{3} \) or \( 2I = \frac{\pi}{3} \Rightarrow I = \frac{\pi}{6} \).

Step 5: Conclusion on Statements

Statement-I is true: the computed value of \( I \) matches the statement's claim. Statement-II is true, and it is the reason Statement-I holds, as it is required to utilize the property to evaluate the integral.

Key Concepts

Integral Calculus

Integral calculus is an essential part of calculus that focuses on the concept of integration. Integration is the process of finding integrals, which are used to calculate areas, volumes, and other quantities that accumulate over an interval. There are two types of integrals: definite and indefinite integrals.

A definite integral is expressed with limits of integration, like in the form \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are the lower and upper limits, respectively. This represents the exact area under the curve \( f(x) \) from \( x=a \) to \( x=b \).

The process of integration is the inverse of differentiation. While differentiation breaks down a function into its rates of change, integration compiles it back into a whole. It plays a crucial role in physics, engineering, and probability, where cumulative quantities need to be determined.

• Definite integrals have bounds and calculate exact values.
• Indefinite integrals represent families of functions without boundaries.

Symmetry Property of Integrals

The symmetry property of integrals is a powerful tool in integral calculus that can simplify the evaluation process of certain integrals. This property states:

• \( \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx \)

This symmetry can often reduce complex integrals into simpler forms by leveraging the symmetric characteristics of the function or the interval of integration.

In the exercise, this property is applied to solve the problem at hand. By setting up the symmetric expression \( \pi/6+\pi/3=\pi/2 \), we can transform the function into one that might be easier or simpler to evaluate, such as transforming \( \tan(x) \) into \( \cot(x) \) by symmetry.

Using this property effectively requires understanding how function transformation relates to integration, which can involve substituting one trigonometric identity for another. This helps simplify the computation, illustrating the utility of symmetry in finding integrals.

Trigonometric Integration

Trigonometric integration involves integrating functions that contain trigonometric functions like \( \sin(x) \), \( \cos(x) \), \( \tan(x) \), and their reciprocals, such as \( \cot(x) \), \( \sec(x) \), and \( \csc(x) \). These types of integrals are common in calculus and often require specific techniques to solve.

Identifying relationships between trigonometric functions is key to successful trigonometric integration. In the original exercise, we specifically employed trigonometric identities:

• \( \tan(\frac{\pi}{2} - x) = \cot(x) \)

This identity was used to transform the function, allowing us to simplify the calculations by transforming a \( \tan(x) \) function into \( \cot(x) \). This is a common strategy in calculus when tackling integration problems that involve trigonometric expressions.

Trigonometric integration is crucial for solving problems in physics, engineering, and other fields where wave patterns and periodic functions play a role. It helps analyze and predict behaviors modeled by sine, cosine, and tangent functions.