Problem 194 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(\int_{0}^{\pi} x f(\sin x) d x=A \int_{0}^{\pi / 2} f(\sin x) d x\), then \(A\) is (A) 0 (B) \(\pi\) (C) \(\frac{\pi}{4}\) (D) \(2 \pi\)

Step-by-Step Solution

Answer

The value of A is \(\pi\), option (B).

Step 1: Analyze the Given Equation

We need to solve \( \int_{0}^{\pi} x f(\sin x) dx = A \int_{0}^{\pi / 2} f(\sin x) dx \) to find the value of \(A\). This requires understanding the structure of both integrals.

Step 2: Simplify the First Integral

Consider the integral \( \int_{0}^{\pi} x f(\sin x) dx \). We can break it down into two integral regions from 0 to \(\frac{\pi}{2}\) and from \(\frac{\pi}{2}\) to \(\pi\), i.e., \( \int_{0}^{\pi/2} x f(\sin x) dx + \int_{\pi/2}^{\pi} x f(\sin x) dx \).

Step 3: Apply Substitution to Simplify

In the second integral, use the substitution \( u = \pi - x \), so \( du = -dx \). The limits change from \(\pi/2\) to \(\pi\) into \(\pi/2\) to 0. Thus, the integral becomes \(-\int_{\pi/2}^{0} (\pi - u) f(\sin(\pi - u)) du = \int_{0}^{\pi/2} (\pi - u) f(\sin u) du\).

Step 4: Combine and Analyze Both Integrals

Combine the two integrals: \( \int_{0}^{\pi/2} x f(\sin x) dx + \int_{0}^{\pi/2} (\pi - x) f(\sin x) dx \). This simplifies to \( \int_{0}^{\pi/2} \pi f(\sin x) dx \) after combining terms.

Step 5: Compare Simplified Integral to Right Side

Equate the simplified left side integral \( \pi \int_{0}^{\pi/2} f(\sin x) dx \) to the right side \( A \int_{0}^{\pi/2} f(\sin x) dx \). Since these two expressions are equal for any function \(f\), the terms inside the integrals cancel out, leaving \(\pi = A\).

Step 6: Conclude the Value of A

From the simplification \( \pi = A \), we can conclude that the value of \(A\) is \(\pi\).

Key Concepts

Integral Calculus

Integral Calculus is a branch of mathematics dealing with integrals, which are the inverse operation of derivatives. It primarily focuses on finding the area under a curve or determining the accumulation of quantities. There are two main types of integrals: definite integrals and indefinite integrals. In this exercise, we're dealing with definite integrals, which calculate the net area under a curve between two specific limits, from a to b. This is denoted as \( \int_{a}^{b} f(x) \, dx \), where \(f(x)\) is the function being integrated.

Definite integrals also include the Fundamental Theorem of Calculus, which connects differentiation and integration. The theorem states that if \(F(x)\) is the antiderivative of \(f(x)\), then:

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

This means the definite integral gives the change in the function value over the interval. In the context of our problem, we calculated the area involving the composition of \(f(\sin(x))\) over two different intervals, [0, \(\pi/2\)] and [\(\pi/2\), \(\pi\)].
Definite integrals are fundamental for solving real-world problems involving rates of change, such as calculating distance, area, volume, and other quantities.

Substitution Method

The substitution method, also known as integration by substitution, is a powerful technique within integral calculus that simplifies the process of finding integrals. It's especially useful when dealing with composite functions or when a direct integration seems challenging. In essence, this method involves changing the variable of integration to make the integral easier to solve.

The general idea is:

• Identify a part of the integrand that can be set as \(u\), such that \(du\) (the derivative of \(u\) with respect to \(x\)) is also within the integral.
• Substitute \(u\) and \(du\) into the integral, transforming it into a simpler form.
• Integrate in terms of \(u\), then substitute back the original variable.

In our exercise, we utilized substitution in the second part of the integral to simplify the interval \([\pi/2, \pi]\). By letting \(u = \pi - x\), we transformed the variable and changed the integration limits from [\(\pi/2\), \(\pi\)] to [\(\pi/2\), 0], which simplifies further calculations.
This substitution rearranges the expression using the identities of trigonometric functions, streamlining the path to find the solution.

Trigonometric Functions

Trigonometric functions are fundamental in mathematics, expressing relationships of angles in triangles. They also extend to periodic functions in advanced calculations like those in calculus problems.

The exercise involves the sine function, \(\sin(x)\), a basic trigonometric function, which describes the y-coordinate of a unit circle at a given angle \(x\). Trigonometric functions have inherent properties allowing simplification using identities, such as \(\sin(\pi - x) = \sin(x)\).

Throughout our problem, \(f(\sin x)\) appears embedded within the integral. The transformation via substitution utilized the identity, simplifying the function to a more accessible form without changing the integral's value:

• \(f(\sin(\pi-x)) = f(\sin x)\)

This symmetry property is crucial for effective integration, especially when trigonometric functions are involved. Understanding these identities and properties helps improve computational efficiency and drives clearer insight into how integrals operate, especially under variable transformations.