Problem 95 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(\int_{0}^{1} \frac{\sin t}{1+t} d t=\alpha\), then the value of the integral \(\int_{4 \pi-2}^{4 \pi} \frac{\sin t / 2}{4 \pi+2-t} d t\) in terms of \(\alpha\) is given by (A) \(2 \alpha\) (B) \(-2 \alpha\) (C) \(\alpha\) (D) \(-\alpha\)

Step-by-Step Solution

Answer

The integral evaluates to \(-\alpha\) (option D).

Step 1: Analyze the given integral

We are given that \( \int_{0}^{1} \frac{\sin t}{1+t} d t = \alpha \). Our task is to find the integral \( \int_{4 \pi-2}^{4 \pi} \frac{\sin \frac{t}{2}}{4 \pi+2-t} d t \) in terms of \( \alpha \).

Step 2: Change of Variables

To relate these integrals, we perform a substitution. Let \( u = 4\pi + 2 - t \). Then \( du = -dt \). When \( t = 4\pi - 2 \), \( u = 4\pi + 2 - (4 \pi - 2) = 4 \). And when \( t = 4\pi \), \( u = 4\pi + 2 - 4\pi = 2 \). Reversing limits, the integral is \(-\int_{2}^{4} \frac{\sin \frac{4\pi + 2 - u}{2}}{u} du \).

Step 3: Simplify the Integral

The expression \( \sin \frac{4\pi + 2 - u}{2} = \sin \left( 2\pi + 1 - \frac{u}{2} \right) = -\sin \left( 1 - \frac{u}{2} \right) \) using the sine addition formula. This results in the integral becoming \( \int_{2}^{4} \frac{\sin \left( \frac{u}{2} - 1 \right)}{u} du \).

Step 4: Use Integral Property

Notice now \( \sin \left( \frac{u}{2} - 1 \right) = \sin \left( 1 - \frac{u}{2} \right) = \sin(x) \), identical to the original integral form but with a different variable and bounds that are changed. Adjust to let \( v = \frac{u}{2} - 1 \), resulting in limits \( 0 \) to \( 1 \).

Step 5: Calculate the Result

With this substitution and recognizing it as equivalent to the original given integral over the considered interval, the integral evaluates to \(-\int_{0}^{1} \frac{\sin v}{1+v} dv = -\alpha\). Therefore, the desired integral is \(-\alpha\).

Key Concepts

Change of Variables

The change of variables, also known as substitution, is a powerful method used in calculus to simplify and solve integrals. In this exercise, we change the variable to make the integral easier to evaluate. By substituting one variable for another, we can potentially transform a complicated integral into a more manageable one. Here's how we did it: for the integral \( \int_{4 \pi-2}^{4 \pi} \frac{\sin \frac{t}{2}}{4 \pi+2-t} \, dt \), we used the substitution \( u = 4\pi + 2 - t \). This substitution effectively reverses the bounds and rearranges the integrand to fit a form that relates to the known integral \( \int_{0}^{1} \frac{\sin t}{1+t} \, dt = \alpha \). The main idea is to create a new variable \( u \) that simplifies integration by modifying the limits and integrand in a predictable way.

Sine Function

The sine function is a fundamental concept in trigonometry, representing the y-coordinate of a point on the unit circle as a function of the angle from the x-axis. It is periodic, with a period of \(2\pi\), meaning it repeats its values every \(2\pi\) radians. In this solution, we take advantage of the properties of the sine function. After substituting \( u \), the expression in the integrand becomes \( \sin \left( \frac{4\pi + 2 - u}{2} \right) \). Using the sine addition formula, we simplify this to \( \sin \left( 2\pi + 1 - \frac{u}{2} \right) \), which is equivalent to \(-\sin \left( 1 - \frac{u}{2} \right) \). The transformation leverages the identity: \( \sin(a + b) = \sin a \cos b + \cos a \sin b \). Since sine is an odd function, \(-\sin(x) = \sin(-x)\), this property is crucial for simplifying the integral.

Integral Properties

Integral properties are rules that help solve and simplify integrals. They include linearity, bounds exchange, and the odd/even function properties. These properties allow the transformation of integrals into more practical forms for calculation. In this exercise, we used two integral properties extensively:

• Reversal of limits: Changing the integration limits from \([4, 2]\) to \([2, 4]\) creates a negative sign, converting the integral to its equivalent opposite.
• Odd function property: Utilized when simplifying \( \sin \) inconsistencies during integration. This means that \(-\sin(x)\) reverses to \(\sin(x)\) upon reflection across the origin.

By employing these properties, we took advantage of the similarities between the original and transformed integrals to ultimately equate one in terms of the known integral \( \alpha \).

JEE Main Mathematics

JEE Main Mathematics examinations often include questions that test deep understanding of fundamental concepts, such as integration and trigonometric functions. The question in this exercise is a typical example, requiring both conceptual clarity and procedural dexterity. Students are expected to understand integral calculus, specifically the use of substitution and integral properties. They should also recognize trigonometric identities and their applications in simplifying expressions. To perform well in JEE Main Mathematics, focus on:

• Practice: Regularly solving a variety of problems helps reinforce key concepts.
• Conceptual Understanding: Grasp the underlying principles rather than memorizing procedures.
• Time Management: Practice solving problems within a time limit to better manage exam pressure.

By mastering these skills, students can better tackle complex integration problems like this one, reflecting the high standards and expectations of JEE Main Mathematics.