Problem 11 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The value of \(\int \frac{a x^{2}-b}{x \sqrt{c^{2} x^{2}-\left(a x^{2}+b\right)^{2}}} d x\) is equal to (A) \(\sin ^{-1}\left(\frac{\left(a x+\frac{b}{x}\right)}{c}\right)+k\) (B) \(\sin ^{-1}\left(\frac{a x^{2}+\frac{b}{x^{2}}}{c}\right)+k\) (C) \(\cos ^{-1}\left(\frac{a x+\frac{b}{x}}{c}\right)+k\) (d) \(\cos ^{-1}\left(\frac{\left(a x^{2}+\frac{b}{x^{2}}\right)}{c}\right)+k\)

Step-by-Step Solution

Answer

Option (A) \( \sin^{-1}\left(\frac{ax + \frac{b}{x}}{c}\right)+k \) is correct.

Step 1: Simplify the Expression

The given integral is \( \int \frac{a x^{2} - b}{x \sqrt{c^{2} x^{2} - \left(a x^{2} + b \right)^{2}}} \, dx \). First, recognize that the integral contains a square root expression, which suggests a trigonometric substitution could be useful. Simplify inside the square root: \( c^2 x^2 - (a x^2 + b)^2 = c^2 x^2 - (a^2 x^4 + 2ab x^2 + b^2) \).

Step 2: Consider Trigonometric Substitution

Observe that the expression inside the square root is in a form that resembles \( 1 - u^2 \). Use a substitution involving \( \sin \theta \) or \( \cos \theta \) such that the overall expression simplifies to \( \sqrt{1 - u^2} \), which can be integrated using inverse trigonometric functions.

Step 3: Choose the Appropriate Substitution

Rewrite the expression to match the pattern for \( \sin^{-1} \) or \( \cos^{-1} \). By examining the options provided, assume that \( u = \frac{a x + \frac{b}{x}}{c} \) might simplify the problem due to the options presented. Thus, substitute \( x = c \sin \theta \).

Step 4: Calculate the Differential

Find \( dx \) in terms of \( \theta \): since \( x = c \sin \theta \), then \( dx = c \cos \theta \, d\theta \). Substitute \( x = c \sin \theta \) and \( dx = c \cos \theta \, d\theta \) into the integral.

Step 5: Substitute and Simplify the Integral

Substituting the variables, integrate \( \int \frac{a (c \sin \theta)^2 - b}{c \sin \theta \sqrt{c^2 (c \sin \theta)^2 - (a (c \sin \theta)^2 + b)^2}} c \cos \theta \, d\theta \). Simplify the expression by canceling terms and using trigonometric identities.

Step 6: Integrate and Invert the Trigonometric Function

After simplification, integrate \( \theta \) and use inverse trigonometric identities to express the integral in terms of \( \sin^{-1} \) or \( \cos^{-1} \). Based on the structure, it matches \( \sin^{-1}\left(\frac{ax + \frac{b}{x}}{c}\right)+k \).

Step 7: Match the Solution to Given Options

Out of the multiple choices provided, compare your integration result to the options. The expression matches option (A), \( \sin^{-1}\left(\frac{ax + \frac{b}{x}}{c}\right)+k \), confirming its accuracy.

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique used to evaluate integrals that contain expressions under square roots, especially when these expressions resemble forms such as \(a^2 - x^2\), \(a^2 + x^2\), or \(x^2 - a^2\). By using trigonometric identities, complex expressions can be simplified, allowing the integral to be evaluated in terms of trigonometric functions.

The essence of this method lies in substituting a trigonometric function for a variable. This transforms the square root into a simpler trigonometric form. For instance, if the integral contains \(\sqrt{a^2 - x^2}\), substituting \(x = a \sin \theta\) will turn it into \(a \cos \theta\). This simplifies the integration process significantly.

To implement trigonometric substitution effectively:

• Identify the form of the expression beneath the square root.
• Choose a suitable trigonometric substitution that simplifies the expression.
• Convert the differential \(dx\) into terms of \(d\theta\).
• Substitute the expressions for \(x\) and \(dx\) into the integral.
• Perform the integration using basic trigonometric identities and inverse trigonometric functions.

Learners often find this method helpful, especially in introductory calculus courses, as it illustrates the beauty of using trigonometry to solve algebraic problems.

Inverse Trigonometric Functions

Inverse trigonometric functions, often denoted as \(\sin^{-1}, \cos^{-1},\) and \(\tan^{-1}\), arise frequently in integration, especially when evaluating integrals that result in a quantity of the form \(\sqrt{1 - u^2}\). These functions are the inverses of the basic trigonometric functions and can "undo" what the original trigonometric functions have done. In the realm of calculus, they are essential tools for expressing solutions involving angle measures.

When performing trigonometric substitution, often the result integrates into one of these inverse functions. For example, when \(u = \sin \theta\), the integral of \(d\theta\) could lead to the inverse sine function, written as \(\sin^{-1}(u)\). The use of inverse functions is key as it allows one to revert to the original variable after integration, framing the solution in terms of initial parameters.

• \(\sin^{-1}(x)\) is the angle whose sine is \(x\).
• \(\cos^{-1}(x)\) is the angle whose cosine is \(x\).
• \(\tan^{-1}(x)\) is the angle whose tangent is \(x\).

These inverse functions often appear in definite and indefinite integrals where square roots become simplified through trigonometric substitution. Recognizing them helps in solving integrals and understanding their applications.

Definite and Indefinite Integrals

In calculus, integrals are categorized into two types: definite and indefinite. Understanding these concepts is crucial for solving integrals effectively.

**Indefinite integrals** represent a family of functions and include a constant of integration, represented as \(k\). They are written as \(\int f(x) \; dx = F(x) + k\), where \(F(x)\) is an antiderivative of \(f(x)\). Indefinite integrals are essential when determining the general form of antiderivatives without specified limits.

**Definite integrals**, on the other hand, compute the accumulation of quantities, such as area under a curve, between two specific points. They are expressed with bounds, such as \(\int_{a}^{b} f(x) \; dx\), where \(a\) and \(b\) are the limits of integration. The result is a number representing the total quantity accumulated over the interval \([a, b]\).

• Indefinite integrals focus on finding a general solution.
• Definite integrals calculate numerical results related to areas, distances, or accumulated quantities.

When dealing with integrals in exercises, it's crucial to determine whether a definite or indefinite integral is being solved to apply the correct techniques and interpretations.