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

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 option is (A) \( I_{2, n+1} - I_{2, n} = I_{1, n} \).

Step 1: Express \( I_{1, n} \)

First, express the integral \( I_{1, n} \):\[I_{1, n} = \int_{0}^{\pi/2} \frac{\sin((2n-1)x)}{\sin x} \, dx.\]Recognize that this is a periodic function and can be expressed as a definite integral that calculates areas under the curve.

Step 2: Express \( I_{2, n} \)

Now, write down the expression for \( I_{2, n} \):\[I_{2, n} = \int_{0}^{\pi/2} \frac{\sin^2(nx)}{\sin^2 x} \, dx.\]This integral evaluates the square of sine terms divided by the square of \( \sin x \) over the given interval.

Step 3: Analyze the Difference

Consider the difference \( I_{2, n+1} - I_{2, n} \):\[I_{2, n+1} = \int_{0}^{\pi/2} \frac{\sin^2((n+1)x)}{\sin^2 x} \, dx\]Subtract \( I_{2, n} \) from \( I_{2, n+1} \):\[I_{2, n+1} - I_{2, n} = \int_{0}^{\pi/2} \left( \frac{\sin^2((n+1)x)}{\sin^2 x} - \frac{\sin^2(nx)}{\sin^2 x} \right) \, dx.\]

Step 4: Simplify and Compare

Now, simplify the expression derived in Step 3 using trigonometric identities and compare it with the given options. Through properties of trigonometric identities such as angle addition formulas, show that:\[I_{2, n+1} - I_{2, n} = I_{1, n}.\]This is because \( \sin^2((n+1)x) - \sin^2(nx) \) can be re-written in terms of sine cosine differences where exactly matches \( I_{1, n} \) after integration.

Key Concepts

Trigonometric Integrals

Trigonometric integrals are a type of integral that involves trigonometric functions, like sine and cosine. These functions are periodic, which means they repeat their values in regular intervals. In the case of the original exercise, we have integrals like \( I_{1, n} = \int_{0}^{\pi/2} \frac{\sin((2n-1)x)}{\sin x} \, dx \) and \( I_{2, n} = \int_{0}^{\pi/2} \frac{\sin^2(nx)}{\sin^2 x} \, dx \). These involve sine ratios within the integral limits of 0 and \( \pi/2 \). This particular setup helps in calculating areas under trigonometric curves.
Exploring these integrals helps us extend our understanding of how trigonometric identities can simplify expressions. For such integrals, employing trigonometric identities allows us to reduce the complexity step by step.

• This reduction often involves using familiar identities like \( \sin^2(x) + \cos^2(x) = 1 \).
• Also, angle addition or subtraction formulas can prove useful.
• Observing these identities at play can simplify evaluating the integral significantly.

Understanding trigonometric integrals, therefore, is not just about computing integrals of sine and cosine functions but also about appreciating the flexibility and transformations brought upon by their inherent periodic properties.

Integration Techniques

Solving integrals often requires sophisticated techniques to simplify and compute them efficiently. Integration techniques are strategies or methods applied to find integrals that are not straightforward to evaluate. Techniques for solving trigonometric integrals like those in the original exercise include methods like substitution, trigonometric identities, and algebraic manipulation.
One useful technique involves substitution, which can simplify complex trigonometric expressions into more manageable ones. In these kinds of scenarios:

• Choose a substitution that simplifies the function or matches a known integral.
• Re-express the integral in terms of the new variable, making it easier to integrate.

Apart from substitution, recognizing patterns using trigonometric identities can dramatically simplify solving. For example, using \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \) or similar transformations can be really helpful.
Through these techniques, integrals that initially seem daunting can become approachable, showing how versatile and robust our toolkit for integration is.

Properties of Definite Integrals

When solving problems involving definite integrals, understanding their fundamental properties is crucial. Definite integrals compute the accumulation of quantities, such as areas under curves, over specific intervals. They provide precise values rather than indefinite integrals, which need a constant of integration. In this original exercise:

• The expression \( I_{2, n+1} - I_{2, n} \) leverages definite integral properties by focusing on specific interval differences.
• Properties like additivity and the linearity of integrals play pivotal roles in breaking down expressions.

The key takeaway with definite integrals is they handle boundary values, turning them into a numerical result. By subtracting one from another, you can determine the change in accumulated quantities over adjustments in the function. Many properties, such as:

• \( \int_a^b [f(x) \pm g(x)] \, dx = \int_a^b f(x) \, dx \pm \int_a^b g(x) \, dx \)
• \( \int_a^b c \cdot f(x) \, dx = c \cdot \int_a^b f(x) \, dx \)

help us manipulate and simplify complex integrals, aiding in obtaining solutions more efficiently. Mastering these properties allows for logical reasoning and systematic problem-solving approaches.