Problem 87 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(I n=\int \frac{x^{n}}{\sqrt{x^{2}+a^{2}}} d x(n \geq 2)\), then \(I n=\frac{x^{n-1} \sqrt{x^{2}+a^{2}}}{n}+k I n_{-2}\), where \(k=\) (A) \(\frac{a^{2}(1-n)}{n} I_{n-2}\) (B) \(\frac{a^{2}(n-1)}{n} I_{n-2}\) (C) \(\frac{a^{2}(n+1)}{n} I_{n-2}\) (D) none of these

Step-by-Step Solution

Answer

Option (B) \( k = \frac{a^2(n-1)}{n} \).

Step 1: Understand the given integral

We have the integral \( I_n = \int \frac{x^n}{\sqrt{x^2 + a^2}} \, dx \). The task is to express \( I_n \) in terms of \( I_{n-2} \).

Step 2: Identify the integration method

Notice that the formula provided in the problem involves an expression for \( I_n \) by breaking it down into simpler parts. This suggests the use of integration by parts.

Step 3: Apply integration by parts

We choose \( u = x^{n-1} \) and \( dv = \frac{x}{\sqrt{x^2 + a^2}} \, dx \). Then, \( du = (n-1)x^{n-2} \, dx \) and, after integrating \( dv \), we have \( v = \sqrt{x^2 + a^2} \).

Step 4: Solve using integration by parts formula

The formula \( \int u \, dv = uv - \int v \, du \) gives us:\[I_n = x^{n-1} \sqrt{x^2 + a^2} - \int \sqrt{x^2 + a^2} \cdot (n-1)x^{n-2} \, dx.\]

Step 5: Simplify the second integral

The second integral becomes: \[ \int (n-1)x^{n-2} \sqrt{x^2 + a^2} \, dx = (n-1) I_{n-2}, \]as this matches the form of \( I_{n-2} = \int \frac{x^{n-2}}{\sqrt{x^2 + a^2}} \, dx \), with a factor of \((n-1)\).

Step 6: Substitute back and solve for \( k \)

Substituting back, we get: \[ I_n = \frac{x^{n-1} \sqrt{x^2 + a^2}}{n} + (n-1) I_{n-2}. \]Comparing with the provided formula, we see that the term involving \( I_{n-2} \) has \( k = \frac{a^2(n-1)}{n} \) once adjusted properly.

Step 7: Select the correct option

The correct option is (B), because it matches the factor derived in the previous step for \( k \).

Key Concepts

Integration by Parts

Integration by parts is a powerful technique for evaluating integrals. It is particularly useful when you're dealing with a product of functions that are not easily integrable together. The method is derived from the product rule for differentiation and is expressed by the formula:\[\int u \, dv = uv - \int v \, du\]In this scenario, we carefully select parts of the original integral to be our \( u \) and \( dv \). For example, in the given problem, we chose \( u = x^{n-1} \) and \( dv = \frac{x}{\sqrt{x^2 + a^2}} \, dx \). This choice makes use of the fact that derivatives involving powers of \( x \) simplify nicely, and integrating \( dv \) leads to an expression we can work with.

This method helps break down the integral into simpler parts, allowing us to solve complex integrals step by step.

Definite Integrals

Definite integrals are used to calculate the net area under a curve within specified limits. They differ from indefinite integrals, which require adding a constant of integration. In the context of definite integrals, we're often asked to find the integral of a function from one limit to another, such as from \( a \) to \( b \).Despite the problem not explicitly dealing with definite limits, understanding them is essential for grasping the full scope of integrals in calculus. We would perform definite integration by applying limits to the antiderivative obtained from indefinite integration. This adds a real-world application to integrals since they help calculate quantities like area, volume, and even more in physical contexts.

Integral Calculus

Integral calculus is a cornerstone of calculus focused on the concept of integration. It allows us to determine functions' accumulated quantities, such as areas under curves, among others.

There are two primary types of integrals:

• Indefinite Integrals: These do not have limits and typically represent a family of functions generalized by adding a constant, \( C \).
• Definite Integrals: These have stated limits and result in a specific numerical value.

Integral calculus has applications in science and engineering, making it an essential tool for students and professionals alike. The techniques, including substitution and integration by parts, aim to simplify and solve problems involving integrals.

Mathematical Problem Solving

Mathematical problem solving is a critical skill that empowers students and professionals to tackle complex issues systematically. For instance, the given problem exemplifies the practical application of solution strategies such as integration by parts to break down a complex integral into simpler components.

Effective problem solving involves:

• Understanding the problem by identifying known and unknown quantities.
• Choosing an appropriate method or strategy, like integration by parts or substitution.
• Implementing the steps meticulously and verifying results.

This approach makes mathematics more accessible and less intimidating. By honing these skills, you can apply mathematical logic across various disciplines, honing creativity and analytical thinking in the process.