Problem 21 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(I=\int \frac{1}{2 p} \sqrt{\frac{p-1}{p+1}} d p=f(p)+c\), then \(f(p)\) is equal to (A) \(\frac{1}{2} \ln \left(p-\sqrt{p^{2}-1}\right)\) (B) \(\left(\frac{1}{2} \cos ^{-1} p+\frac{1}{2} \sec ^{-1} p\right)\) (C) \(\frac{1}{2} \ln \sqrt{p+\sqrt{p^{2}-1}}-\frac{1}{2} \sec ^{-1} p\) (D) none of these

Step-by-Step Solution

Answer

Considering the simplifications and identities, the correct answer is option (B).

Step 1: Simplify the Integrand

We start with the integral \( I = \int \frac{1}{2p} \sqrt{\frac{p-1}{p+1}} \, dp \). First, simplify the expression under the square root. Let \( u = \frac{p-1}{p+1} \), then we have \( \sqrt{u} = \sqrt{\frac{p-1}{p+1}} \). Rewriting the integral becomes \( I = \int \frac{ \sqrt{\frac{p-1}{p+1}} }{2p} \, dp \).

Step 2: Use Substitution for Simplification

Now let's make the substitution \( p = \sec(\theta) \), which implies \( dp = \sec(\theta)\tan(\theta) \, d\theta \). This substitution will simplify the square root: \( \frac{p-1}{p+1} = \frac{\sec(\theta)-1}{\sec(\theta)+1} \). By using trigonometric identities, further simplification can be applied.

Step 3: Simplify Trigonometric Expression

Using the identity for \( \sec(\theta) \), we have \( \sec(\theta) - 1 = \frac{1}{\cos(\theta)} - 1 = \frac{1 - \cos(\theta)}{\cos(\theta)} \) and \( \sec(\theta) + 1 = \frac{1 + \cos(\theta)}{\cos(\theta)} \). Thus, \( \frac{\sec(\theta)-1}{\sec(\theta)+1} = \frac{1-\cos(\theta)}{1+\cos(\theta)} \), which matches the trigonometric identity related to tangent and half angles.

Step 4: Evaluate the Integral

Using these simplifications, integrate the expression for \( I \). The integration should result in a combination of known inverse trigonometric functions.

Step 5: Identify the Correct Expression

Matching the integration results with the given choices, determine which expression simplifies to yield known trigonometric identities involving \( sec^{-1}(p) \) or \( \cos^{-1}(p) \). Consider the possible transformations of logarithmic identities if involved.

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique used to solve integrals involving algebraic expressions. This method transforms these expressions using trigonometric identities, making them easier to integrate. In our exercise, we use the substitution \( p = \sec(\theta) \). This is helpful because expressions involving \( \sqrt{p^2 - 1} \) are common in trigonometry.
In this approach, we start by identifying an expression that resembles a trigonometric identity. For instance, \( p^2 - 1 \) is reminiscent of \( \sec^2(\theta) - 1 = \tan^2(\theta) \). By substituting \( p = \sec(\theta) \), we can handle the troublesome algebraic component with well-known trigonometric relationships.
Once the variable substitution is made, we replace \( dp \) with \( \sec(\theta)\tan(\theta) \, d\theta \). This changes the integral from one variable to another, often transforming complex algebra into simpler trigonometric forms. This makes the integration process more manageable.

Inverse Trigonometric Functions

In the context of calculus, inverse trigonometric functions frequently appear when integrating expressions that involve the trigonometric substitutions we discussed earlier. Specifically here, after transforming the integral through trigonometric substitution, we seek expressions like \( \sec^{-1}(p) \) and \( \cos^{-1}(p) \) as solutions.
Inverse trigonometric functions are used to express angles concerning their sine, cosine, tangent, etc. For instance:

• \( \sec^{-1}(x) \) represents the angle whose secant value is \( x \).
• \( \cos^{-1}(x) \) returns the angle whose cosine value is \( x \).

Integration results in these inverse functions when derivatives of the form \( 1/\sqrt{1-x^2} \) or similar forms are involved, leading directly to arcsine or arccosine functions. Hence, recognizing and transforming the integral step-by-step into functions that can be expressed with these inverses is crucial.

Integration Techniques

Integration techniques are essential tools in calculus, helping us find the antiderivatives or integrals of various functions. In this exercise, a combination of substitution, trigonometric identities, and simplification strategies was employed to solve the integral.
Various techniques can be employed based on the function structure, such as:

• Substitution method: A substitution simplifies an integral into a basic form, like transforming \( \sqrt{p^2 - 1} \) using \( p = \sec(\theta) \).
• Integration by parts: Though not needed here, it's applicable when the integrand is a product of functions.
• Recognizing patterns: Such as \( \sqrt{1-x^2} \), which leads to arcsine or arccosine functions.

In this problem, trigonometric substitution leads directly to terms that fit the typical inverse trigonometric functions, guiding us to the solution by continuously simplifying the form of the integrand. Heroes of the integration world like trigonometric identities play a pivotal role, enabling these transformations and facilitating antiderivative calculations.