Problem 28 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If the integral \(\int \frac{5 \tan x}{\tan x-2} d x=x+a \ln |\sin x-2 \cos x|\) \(+k\), then \(a\) is equal to: (A) \(-1\) (B) \(-2\) (C) 1 (D) 2

Step-by-Step Solution

Answer

\(a = 2\) (Option D).

Step 1: Identify the Integral Structure

The given integral is \( \int \frac{5 \tan x}{\tan x - 2} \, dx \). We are asked to find the value of \( a \) such that the integral equates to \( x + a \ln |\sin x - 2 \cos x| + k \).

Step 2: Simplify the Integral Expression

Re-write \( 5 \tan x \) as \( 5 \frac{\sin x}{\cos x} \) and \( \tan x - 2 \) as \( \frac{\sin x - 2 \cos x}{\cos x} \). Thus, the integral becomes \( \int 5 \frac{\sin x}{\frac{\sin x - 2 \cos x}{\cos x}} \, dx = \int \frac{5 \sin x \cos x}{\sin x - 2 \cos x} \, dx \).

Step 3: Set up a Suitable Substitution

Use substitution to simplify the integral. Let \( u = \sin x - 2 \cos x \), then \( du = \cos x \, dx + 2 \sin x \, dx \). Modify it to express \( \cos x \, dx \) as \( du = \cos x(1 + 2\tan x) \, dx = (\cos x + 2 \sin x) \, dx \).

Step 4: Rewrite the Integral with Substitution

Given the substitution, express \( 5 \sin x \cos x \, dx \) in terms of \( u \). Notice that \( 5 \sin x \cos x \, dx = \frac{5}{2} du \) because \( \sin x - 2 \cos x = u \) implies \( 5 \cos x \sin x = 2.5 \times 2 \sin x \cos x \), so we arrive at \( \int \frac{5}{u} \, \frac{du}{2.5} = 2 \int \frac{1}{u} \, du \).

Step 5: Integrate and Solve for \( a \)

Evaluate the integral: \( 2 \int \frac{1}{u} \, du = 2 \ln |u| + C = 2 \ln |\sin x - 2 \cos x| + C \). Compare this with the original expression \( x + a \ln |\sin x - 2 \cos x| + k \). The coefficient \( a \) must be equal to \( 2 \).

Step 6: Verify the Answer

Since \( a = 2 \), confirm that the substitution process aligns with the original integral setup, ensuring the logical steps follow correctly. Hence, \( a \) is \( 2 \).

Key Concepts

Substitution Method

The substitution method is a powerful technique in integral calculus designed to simplify integrals by changing the variable of integration. In simpler terms, it helps us transform a complex integral into a simpler one that is easier to evaluate.

Imagine the substitution method as changing the perspective of the problem. It allows us to "substitute" a part of the integral with a new variable. This new variable often simplifies the integral into a basic form we know how to solve.

• Identify an expression within the integral that makes the problem complex.
• Substitute this expression with a new variable, usually denoted as \( u \).
• Calculate the derivative \( du \) of this new variable \( u \) with respect to \( x \), and express \( dx \) in terms of \( du \).
• Rewrite the entire integral in terms of \( u \), making it simpler and more straightforward to evaluate.

In our original exercise, we substituted \( u = \sin x - 2 \cos x \), which turned the complicated trigonometric integral into a simpler logarithmic form. This made the evaluation much more manageable!

Definite Integrals

Definite integrals are a fundamental concept in integral calculus, representing the area under a curve within specific limits. Unlike indefinite integrals, which produce a family of functions plus a constant, definite integrals yield a specific numerical value.

Think of definite integrals as calculating the exact area "bounded" by the curve of a function between two endpoints on the x-axis. This involves:

• Calculating the antiderivative of the function, also known as the indefinite integral.
• Applying the fundamental theorem of calculus, which uses the antiderivative to evaluate the integral across the interval \( [a, b] \).
• Subtracting the value of the antiderivative at the lower bound \( a \) from the value at the upper bound \( b \).

In this exercise, even though we're not calculating a definite integral, understanding these concepts helps frame our thought process in evaluating the involved integrals from a limits-based perspective.

Trigonometric Integrals

Trigonometric integrals involve functions that include trigonometric expressions such as sine, cosine, and tangent. These types of integrals often occur when evaluating problems dealing with periodic phenomena or wave-like behavior.

When tackling trigonometric integrals, we often use techniques like substitution to transform the integral into a more conventional form:

• Identify the trigonometric identities present within the integral, such as \( \sin^2 x + \cos^2 x = 1 \).
• Apply appropriate substitutions or transformations to simplify the trigonometric expressions.
• Use known integral results or derivations involving basic trigonometric functions to solve the integral.

In our exercise, we engaged with trigonometric functions by expressing \( \tan x \) in terms of \( \sin x \) and \( \cos x \). The substitution process then effectively converted it into a form that resembled an easier, known integral result, allowing us to solve it.