Problem 144 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(A_{n}=\int_{0}^{\pi / 2} \frac{\sin (2 n-1) x}{\sin x} d x ;\) \(B_{n}=\int_{0}^{\pi / 2}\left(\frac{\sin n x}{\sin x}\right)^{2} d x ;\) for \(n \in N\), then (A) \(A_{n+1}=A_{n}\) (B) \(B_{n+1}=B_{n}\) (C) \(A_{n+1}-A_{n}=B_{n+1}\) (D) \(B_{n+1}-B_{n}=A_{n+1}\)

Step-by-Step Solution

Answer

Option (A) is correct.

Step 1: Evaluate An

The integral for \( A_n \) is given by \( A_{n} = \int_{0}^{\pi / 2} \frac{\sin (2n-1)x}{\sin x} \, dx \). By using the trigonometric identity \( \sin(2n-1)x = \sin((2n-2)x + x) = \sin(2n-2)x \cos x + \cos(2n-2)x \sin x \), and then applying the properties of integrals combined with symmetry, it can be shown that \( A_{n+1} = A_n \). Thus, option (A) is correct.

Step 2: Evaluate Bn

The integral for \( B_n \) is given by \( B_{n} = \int_{0}^{\pi/2} \left( \frac{\sin nx}{\sin x} \right)^2 \, dx \). This term arises from the Parseval's identity where \( B_{n+1} = B_n \) and independent evaluations might not hold similarly to \( A_n \). So, option (B) needs further insights.

Step 3: Check C Equation

The equation \( A_{n+1} - A_n = B_{n+1} \) would not hold because both \( A_{n+1} \) and \( A_n \) are equal, implying \( A_{n+1} - A_n = 0 \). Therefore, option (C) is incorrect.

Step 4: Check D Equation

Considering options and their relationships derived from Parseval's identities and trigonometry, the noteworthy result is that neither \( A_n \) nor \( B_n \) straightforwardly gives \( B_{n+1} - B_n = A_{n+1} \). Hence, option (D) is also incorrect with the earlier reviews.

Key Concepts

Trigonometric Identities

Trigonometric identities are fundamental tools in mathematics, especially in calculus and analytical trigonometry. They allow us to simplify complex trigonometric expressions and make them easier to work with.
These identities include relations like:

• \( \sin(A + B) = \sin A \cos B + \cos A \sin B \)
• \( \cos(A + B) = \cos A \cos B - \sin A \sin B \)
• \( \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \)

In this exercise's context, using trigonometric identities like \( \sin(2n-1)x = \sin((2n-2)x + x) \), helps to decompose and navigate the integral expressions. This helps find symmetries or eliminate variables, simplifying the problem immensely.
Grasping these identities not only aids in solving integrals but also gives a deeper understanding of angles and geometric relations. Often, these identities are used to reformulate integral expressions, making integration processes much simpler and more straightforward.

Parseval's Identity

Parseval's identity is an important concept in mathematics, particularly in the context of Fourier analysis. This identity extends the Pythagorean theorem to functions, expressing that the sum of the square of a sequence is equal to the integral of the square of the function, often seen in the realm of periodic functions.
In terms of integration, Parseval's identity can connect sums of squares of coefficients in series with integrals of squared functions. This is often applied when dealing with Fourier series, which represent functions as sums of sines and cosines.
Applying Parseval’s identity to the exercise, the term \( B_n \) relates to these concepts and reinforces the fact that relationships between integrals can exhibit similar properties to series sums. Recognizing this connection allows mathematicians to apply known results from series to integrals, providing valuable insights in simplifying and understanding complex integral evaluations.

Symmetry in Integration

Symmetry in integration can often simplify solving definite integrals, especially when the functions involved possess symmetrical properties. If a function is symmetric around an axis or a point, it may reduce the complexity of evaluating the integral over a symmetrical interval.
In essence, symmetry allows for a reduction in effort by reducing the range of integration or indicating that portions of the integral can cancel out.

• If a function is even (symmetrical around the y-axis), \( \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx \).
• If a function is odd, the integral across symmetric limits results in zero, \( \int_{-a}^{a} f(x) \, dx = 0 \).

In the context of the problem, symmetry plays a crucial role as it allows the integral \( A_n \) to remain constant across increments of \( n \). Identifying and exploiting such symmetries can largely ease the evaluation process, demonstrating that understanding the nature of symmetry can be incredibly powerful in mathematical analysis.