Question

The value of the integral \(\int_{-\pi / 2}^{\pi / 2} \log \left(\frac{a-\sin \theta}{a+\sin \theta}\right) d \theta\), \(a>1\) is - (1) 0 (2) 1 (3) 2 (4) \(-2\)

Short answer

0

Step-by-step solution

Simplify the Integral

Consider the integral \(\textrm{I} = \int_{-\pi / 2}^{\pi / 2} \log(\frac{a - \sin \theta}{a + \sin \theta}) d \theta\). Simplify it by separating it into two integrals.

Use Symmetry of the Integrand

Notice that the function \(\textrm{f}(\theta) = \log(\frac{a - \sin \theta}{a + \sin \theta})\) is odd because \(\textrm{f}(-\theta) = -\textrm{f}(\theta)\).

Evaluate the Integral of an Odd Function

Since the integrand is odd and the interval \([-\pi / 2, \pi / 2]\) is symmetric around zero, the integral of an odd function over a symmetric interval is zero, i.e., \(\textrm{I} = 0\).

Key concepts

odd functions

An odd function is a function that satisfies the property that \(f(-x) = -f(x)\). In other words, the graph of an odd function is symmetric with respect to the origin.

For example, in the given integral, the function \(f(\theta) = \log\bigg(\frac{a - \sin \theta}{a + \sin \theta}\bigg)\) is odd. This is because if you substitute \( -\theta \) into the function, you get:
\f(-\theta) = \log \bigg(\frac{a - \sin (-\theta)}{a + \sin (-\theta)}\bigg).
Using the property of sine that \sin(-\theta) = -\sin(\theta)\, this becomes:
\f(-\theta) = \log\bigg(\frac{a + \sin \theta}{a - \sin \theta}\bigg).
This can be rewritten as: \- \log \bigg(\frac{a - \sin \theta}{a + \sin \theta}\bigg).
Also note: -f(\theta) = -\log \bigg(\frac{a - \sin \theta}{a + \sin \theta}\bigg). Hence, \(f(-\theta) = -f(\theta)\), proving that the function is odd.

integral properties

Integrating an odd function over a symmetric interval around zero has a significant property. Specifically, the integral of an odd function over an interval \([-a, a]\) is zero.

Taking our problem as an example, we are looking at the integral: \I = \int_{-\pi / 2}^{\pi / 2} \log \bigg(\frac{a - \sin \theta}{a + \sin \theta}\bigg) d \theta \.

Given the odd nature of \(\log \bigg(\frac{a - \sin \theta}{a + \sin \theta}\bigg)\), and considering that the interval from \(-\pi / 2\) to \(\pi / 2\) is symmetric about zero, we can conclude that this integral evaluates to zero.

This property of integrals is quite helpful in simplifying calculations and determining the results of integrals quickly. When dealing with real-world problems, recognizing the symmetry and properties of functions can save time and effort.

logarithmic functions

Logarithmic functions are functions that involve the logarithm, which is the inverse operation to exponentiation. The logarithm of a number is the exponent by which another fixed number, the base, must be raised to produce that number.

In this problem, we deal with a logarithmic function of the form \(\log \bigg(\frac{a - \sin \theta}{a + \sin \theta}\bigg)\). Here's a breakdown of how to handle such functions:

• The denominator \(a + \sin \theta\) ensures the argument of the logarithm is always positive if \(a > 1\). This maintains the definition domain of the logarithmic function.
• The symmetry and properties of trigonometric functions (like \(\sin \theta\)) play a major role in simplifying logarithmic expressions.

Understanding these logarithmic properties becomes easier when you recognize how they interact with other functions, particularly trigonometric ones in this exercise. The logarithmic function’s core property of transforming multiplication into addition and division into subtraction often simplifies complex expressions and integrals.