Problem 83 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(I_{1, \mathrm{n}}=\int_{0}^{\pi / 2} \frac{\sin (2 n-1) x}{\sin x} d x\) and \(I_{2, \mathrm{n}}=\int_{0}^{\pi / 2} \frac{\sin ^{2} n x}{\sin ^{2} x} d x\), \(n \in N\), then (A) \(I_{2, n+1}-I_{2, n}=I_{1, n}\) (B) \(I_{2, n+1}-I_{2, n}=I_{1, n+1}\) (C) \(I_{2, n+1}+I_{1, n}=I_{2, n}\) (D) \(I_{2, n+1}+I_{1, n+1}=I_{2, n}\)

Step-by-Step Solution

Answer

The correct answer is (B) \(I_{2, n+1}-I_{2, n}=I_{1, n+1}\).

Step 1: Understand the given integrals

We are given two integrals, \(I_{1,n}\) and \(I_{2,n}\), as follows.\[I_{1, n} = \int_{0}^{\pi / 2} \frac{\sin((2n-1)x)}{\sin x} \, dx\] \[I_{2, n} = \int_{0}^{\pi / 2} \frac{\sin^2(nx)}{\sin^2 x} \, dx\]. We need to find a relationship between \(I_{1,n}\) and \(I_{2,n}\) for consecutive values of \(n\).

Step 2: Change of variables in \(I_{2,n}\)

To explore \(I_{2, n+1} - I_{2, n}\), we need to consider the integrals for both \(n\) and \(n+1\). Thus, we view these changes through potential simplifications using identities such as \(\sin^2(A) - \sin^2(B) = \sin(A+B)\sin(A-B)\).

Step 3: Calculate \(I_{2, n+1} - I_{2, n}\)

Expanded, \(I_{2, n+1} = \int_{0}^{\pi / 2} \frac{\sin^2((n+1)x)}{\sin^2 x} \, dx \) and \(I_{2, n} = \int_{0}^{\pi / 2} \frac{\sin^2(nx)}{\sin^2 x} \, dx\). The integrand simplifies to: \[\frac{\sin^2((n+1)x) - \sin^2(nx)}{\sin^2 x} = \frac{\sin((2n+1)x)\sin x}{\sin^2 x} = \frac{\sin((2n+1)x)}{\sin x}\].

Step 4: Deduce that \(I_{2, n+1} - I_{2, n} = I_{1, n+1}\)

By comparison with \(I_{1, n} = \int_{0}^{\pi / 2} \frac{\sin((2n-1)x)}{\sin x} \, dx\), we see that the derived result matches the form of \(I_{1, n+1}\) since the sine function within the integral has the necessary transformation to match \((2n+1)x\). Thus, \(I_{2, n+1} - I_{2, n} = I_{1, n+1}\).

Step 5: Concluding which option fits

With the derivation completed, the relationship \(I_{2, n+1} - I_{2, n} = I_{1, n+1}\) corresponds to option (B). Thus, the answer to the problem is determined.

Key Concepts

Definite Integrals

Definite integrals are an essential part of calculus used to calculate the area under a curve within a specified interval. When we examine a function over a range, say from \( a \) to \( b \), the definite integral \( \int_a^b f(x) \, dx \) allows us to determine the total accumulation of that function.

• The fundamental idea behind definite integrals is the summation of tiny parts, or infinitesimals, that together form the area under a typical curve.
• Unlike indefinite integrals, definite integrals yield a numeric result rather than a general function.

When dealing with problems, set precise limits (here \(0\) to \(\frac{\pi}{2}\)). This provides a specific solution range.
In this exercise, the integrals \( I_{1, n} \) and \( I_{2, n} \) are evaluated over an interval, establishing a baseline to find relationships between these integral expressions. This mathematical process is fundamental in many applications, such as physics and engineering, where it is crucial to evaluate quantities precisely over defined ranges.

Trigonometric Integrals

Trigonometric integrals are a class of integrals involving trigonometric functions like sin, cos, and tan which are recurrent in calculus problems. Understanding how to handle these integrals involves the use of trigonometric identities and substitutions. For example:

• Identities such as \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \) can simplify integrations.
• Substituting variables or using identities helps transform complex trigonometric expressions into simpler forms that are easier to integrate.

In the original exercise, the problem incorporates sine functions, which can be tricky without proper simplification.
For integrals like \( I_{1, n} \) and \( I_{2, n} \), utilizing identities allows the expression to be transformed into a more manageable form (e.g., transforming \( \sin^2(nx) \) into terms of \( \cos \)). These transformations are key to unlocking the simpler integral relationships that hold in trigonometric calculus. By dissecting each term and identifying known identities, complex scenarios can be made tractable, supporting more straightforward resolutions.

Calculus Problem-Solving

Effective calculus problem-solving requires a strategic approach, notably when dealing with integrals. It's not just about computation, but also understanding the theory and relationships.

• First, deeply analyze the problem to understand what's being asked. Break it into smaller, manageable pieces.
• Identify potential simplifications, like using trigonometric identities to reduce complexity.
• Calculate step-by-step, keeping an eye on the big picture for the correct integral relations or transformations.

In the original problem, starting with interpreting the integral expressions \( I_{1, n} \) and \( I_{2, n} \) set the stage for finding links between them. Revisiting each integral's formulation by conceptualizing change, simplification using properties, or contrasting terms becomes the key strategy.
Eventually, determining that \( I_{2, n+1} - I_{2, n} = I_{1, n+1} \) involves critical thinking, transforming ideas, and linking them within the bounds of the defined range. Such rigorous problem-solving not only abides by mathematical conventions but strengthens analytical prowess in solving complex calculus scenarios.