Problem 181 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\int_{-\pi}^{\pi} \frac{2 x(1+\sin x)}{1+\cos ^{2} x} d x\) is (A) \(\frac{\pi^{2}}{4}\) (B) \(\pi^{2}\) (C) zero (D) \(\frac{\pi}{2}\)

Step-by-Step Solution

Answer

(C) zero.

Step 1: Check For Symmetry

The integral limits are symmetric about the origin (-π to π). Thus, we need to check whether the integrand is an odd or even function. Let \( f(x) = \frac{2x(1 + \sin x)}{1 + \cos^2 x} \). To determine the symmetry, calculate \( f(-x) \).

Step 2: Calculate f(-x)

Substitute \(-x\) into the function: \( f(-x) = \frac{2(-x)(1 + \sin (-x))}{1 + \cos^2 (-x)} \).Since \( \sin(-x) = -\sin(x) \) and \( \cos(-x) = \cos(x) \), this simplifies to: \( f(-x) = \frac{-2x(1 - \sin x)}{1 + \cos^2 x} \). Thus, \( f(-x) = -f(x) \), confirming \( f(x) \) is an odd function.

Step 3: Use the Property of Odd Functions

For integrals of odd functions over symmetric limits, \[ \int_{-a}^{a} f(x) \, dx = 0 \]. Since \( f(x) \) is odd and the limits are \(-\pi\) to \(\pi\), the integral evaluates to 0.

Key Concepts

Symmetry of Functions

Symmetry in mathematical functions is a powerful tool in solving integrals, particularly when the limits are symmetric around the origin, like from \(-\pi\) to \(\pi\). Recognizing symmetry can simplify calculations drastically. There are two main types of symmetry:

• Even Symmetry: A function \(f(x)\) is even if \(f(-x) = f(x)\). Graphically, this means the function is symmetric about the y-axis.
• Odd Symmetry: A function \(f(x)\) is odd if \(f(-x) = -f(x)\). This results in the function being symmetric about the origin.

Understanding these symmetries can dramatically alter how we evaluate definite integrals over symmetric limits. It is essential to first determine the type of symmetry involved in the problem, which directly influences the outcome of the integral.

Odd and Even Functions

When dealing with definite integrals, knowing whether a function is odd or even aids in simplification. Let's explore these two cases further:

• Even Functions: Suppose \(f(x)\) is an even function. When integrating over symmetric limits, the result is twice the integral from 0 to \(a\). For example: \(\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx\).
• Odd Functions: If \(f(x)\) is odd and the integral limits are symmetric around zero, the definite integral evaluates to zero: \(\int_{-a}^{a} f(x) \, dx = 0\). This property is extremely useful because it nullifies the integral computation for odd functions over symmetric intervals.

Utilizing these properties can simplify integral calculations, especially when tackling complex functions like the given exercise, where checking for oddness allowed us to conclude the integral result promptly.

Integral Properties

Definite integrals have numerous properties that facilitate the evaluation of integrals, particularly over symmetric limits. Here are some pivotal properties:

• Linearity: The integral of the sum of functions is equal to the sum of the integrals of the functions. Similarly, the integral of a constant multiplied by a function is the constant multiplied by the integral of the function: \(\int_{a}^{b} [c \cdot f(x)] \, dx = c \cdot \int_{a}^{b} f(x) \, dx\).
• Symmetric Limits: For periodic functions, the integral over a full period can sometimes simplify using the function's periodicity and symmetry.
• Zero Result for Odd Functions: As we saw in the exercise, when an odd function is integrated over symmetric limits, the result is zero: \(\int_{-a}^{a} f(x) \, dx = 0\).

By leveraging these properties, you can efficiently tackle definite integrals. In the exercise provided, identifying the function as odd immediately led us to determine the integral's value without further computation.