Problem 139 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\alpha\) is a parameter independent of \(x\) and \(\alpha \neq\) \((2 n+1) \pi, n \in Z\), then the value of the integral \(\int_{0}^{1} \frac{x^{\cos \alpha}-1}{\ln x} d x, x>0, x \neq 1\) is (A) \(\ln (1+\cos \alpha)\) (B) \(\ln (1-\cos \alpha)\) (C) \(\ln |\cos \alpha|\) (D) None of these

Step-by-Step Solution

Answer

The value of the integral is (C) \( \ln |\cos \alpha| \).

Step 1: Identify the Structure of the Integral

The given integral is \( \int_{0}^{1} \frac{x^{\cos \alpha}-1}{\ln x} \, dx \). Observe that the integrand \( \frac{x^{\cos \alpha}-1}{\ln x} \) resembles the form of a derivative of a function. Specifically, this resembles an expression related to differentiating powers or exponentials.

Step 2: Use a Substitution

To simplify the integral, consider the substitution \( x = e^{-t} \), which leads to \( dx = -e^{-t} dt \). This changes the limits of integration as well: when \( x = 0, t = \infty \) and when \( x = 1, t = 0 \). The integral transforms to \(-\int_{\infty}^{0} \frac{(e^{-t})^{\cos \alpha} - 1}{-t} \cdot e^{-t} \, dt \). Simplifying gives \( \int_{0}^{\infty} \frac{e^{-t \cos \alpha} - 1}{t} \, dt \).

Step 3: Recognize a Known Integral

The integral \( \int_{0}^{\infty} \frac{e^{-t \cos \alpha} - 1}{t} \, dt \) is a known result which simplifies to \( \ln |\cos \alpha| \) for \( \alpha eq (2n + 1)\pi, n \in \mathbb{Z} \). This is a standard result in integration involving logarithmic expressions.

Step 4: Determine the Correct Answer from the Options

Comparing the evaluated result from Step 3 with the given options, we find that it matches option (C), \( \ln |\cos \alpha| \). Therefore, the correct answer to this integral problem is \( \ln |\cos \alpha| \).

Key Concepts

Integration Techniques

Integration techniques are methods used to solve integrals, which are essential in calculus for finding areas under curves and other applications. Many techniques exist, each suited to different types of integrals. Key approaches include:

• Substitution: Used to simplify integrals by changing variables.
• Integration by Parts: Useful for integrating the product of two functions.
• Partial Fractions: Breaks down complex rational expressions into simpler fractions.

In our specific exercise, substitution is a key technique. It helps transform the integrand into a more workable form, making the integral easier to evaluate. By recognizing the structure of the integral and knowing which method to apply, solving integrals becomes more manageable.

Substitution Method

The substitution method is a powerful integration technique that involves changing variables to simplify the integral. The idea is to choose a new variable that makes the integral easier to handle. In the given problem, we perform the substitution:

• Letting \( x = e^{-t} \), which implies that \( dx = -e^{-t} dt \).
• This changes the limits of integration: when \( x = 0 \), \( t = fty \), and when \( x = 1 \), \( t = 0 \).

This substitution transforms the integral from \( \int_{0}^{1} \frac{x^{\cos \alpha}-1}{\ln x} \, dx \) to the integral \( \int_{0}^{\infty} \frac{e^{-t \cos \alpha} - 1}{t} \, dt \).
This form is much more straightforward to evaluate. It illustrates how substitution can drastically simplify a seemingly complicated integral. The key is to recognize a pattern that makes implementation of substitution possible.

Logarithmic Expressions

Logarithmic expressions often appear in integration, especially when dealing with exponential functions or scenarios requiring a change of variables. In our example:

• We encounter the logarithmic expression \( \ln x \) in the original integrand.
• The substitution method transforms the problem into dealing with \( \frac{1}{t} \), reflecting conditions where logarithms naturally occur.

In the context of definite integrals, recognizing such patterns can lead to identifying known integral solutions. In the problem, the integral \( \int_{0}^{\infty} \frac{e^{-t \cos \alpha} - 1}{t} \, dt \) is recognized as a standard integral with the result \( \ln |\cos \alpha| \).
Understanding logarithms and their properties helps in solving these integrals efficiently and can lead to swift and correct application of known results.