Problem 6 - A Complete Resource Book in Mathematics for JEE Main

Question

\(\int x\left[f\left(x^{2}\right) g^{\prime \prime}\left(x^{2}\right)-f^{\prime \prime}\left(x^{2}\right) g\left(x^{2}\right)\right] d x\) (A) \(f\left(x^{2}\right) g^{\prime}\left(x^{2}\right)-g\left(x^{2}\right) f^{\prime}\left(x^{2}\right)+c\) (B) \(\frac{1}{2}\left[f\left(x^{2}\right) g\left(x^{2}\right) f^{\prime}\left(x^{2}\right)\right]+c\) (C) \(\frac{1}{2}\left[f\left(x^{2}\right) g^{\prime}\left(x^{2}\right)-g\left(x^{2}\right) f^{\prime}\left(x^{2}\right)\right]+c\) (D) none of the above

Step-by-Step Solution

Answer

The answer is (C).

Step 1: Recognize Substitution

The integral is of the form \( \int x [f(x^2)g''(x^2) - f''(x^2)g(x^2)] \, dx \). Notice that \(x^2\) is a common expression, suggesting a substitution strategy. Let \( u = x^2 \), then \( du = 2x \, dx \). Therefore, \( x \, dx = \frac{1}{2} du \).

Step 2: Apply Substitution

Substitute \(x^2\) with \(u\) and \(x \, dx\) with \(\frac{1}{2} du\) in the integral. The integral becomes: \[ \int \frac{1}{2} [f(u) g''(u) - f''(u) g(u)] \, du \] After multiplying by \(\frac{1}{2}\), it becomes: \[ \frac{1}{2} \int [f(u) g''(u) - f''(u) g(u)] \, du \]

Step 3: Identify a Known Structure

Recognize that the integrand \(f(u) g''(u) - f''(u) g(u)\) is structured like the derivative of a product. Specifically, the expression resembles the result of differentiating \(f(u)g'(u) - g(u)f'(u)\).

Step 4: Integrate By Observation

From the observation in Step 3, the integral \( \int [f(u) g''(u) - f''(u) g(u)] \, du \) aligns with \(f(u)g'(u) - g(u)f'(u)\) due to the derivative relationship. Therefore, \[ \frac{1}{2} [f(u)g'(u) - g(u)f'(u)] + C \] where \(C\) is the constant of integration.

Step 5: Back Substitute

Replace \(u\) with \(x^2\) to revert back to the variable \(x\). Thus, the integral becomes: \[ \frac{1}{2} [f(x^2)g'(x^2) - g(x^2)f'(x^2)] + C \]

Step 6: Match to Given Options

Analyze the solutions provided in the options: (A), (B), (C), and (D). The result from our integration \(\frac{1}{2} [f(x^2)g'(x^2) - g(x^2)f'(x^2)] + C\) matches option (C).

Key Concepts

Substitution Method

The substitution method is a technique used to simplify integrals by changing variables. It's particularly useful when you notice a repeating pattern in your integral, allowing you to streamline the process. In this exercise, the important step was to recognize that \( x^2 \) appears frequently. By letting \( u = x^2 \), we can transform and simplify the integration. When you substitute \( u = x^2 \), it follows that \( du = 2x \, dx \), so \( x \, dx = \frac{1}{2} du \). This transformation changes the integral into a function of \( u \), making it easier to handle.
Understanding and implementing the substitution method involves:

• Identifying a part of the integrand to represent as \( u \).
• Expressing \( du \) in terms of \( dx \) to facilitate a variable transformation.
• Substituting back into the function to simplify and solve the integral.

In this case, substitution was effective because it exploited the repetitive nature of \( x^2 \), resulting in a form that allowed for further manipulation of the integral.

Integration by Parts

Integration by parts is another crucial technique used to simplify and solve certain integrals, based on the product rule for differentiation. The formula is given by \[ \int u \, dv = uv - \int v \, du \]where you choose parts of your integrand to represent \( u \) and \( dv \). This tool is particularly handy when dealing with products of functions, such as polynomials and exponential functions.
Though this specific exercise doesn’t directly use integration by parts, understanding its relevance is essential. Essentially, it helps in recognizing structures that seem too complex at first. The identification of patterns in derivatives often becomes apparent once you have grasped integration by parts principles.

Derivative of a Product

The derivative of a product involves using the product rule that states if \( f \) and \( g \) are functions of \( x \), then \[ (fg)' = f g' + g f' \].In the context of integration, or more specifically in this exercise, recognizing derivatives formed by functions allows us to see through complex expressions. In step 3 of the solution, the observed expression \( f(u)g''(u) - f''(u)g(u) \) was structured similarly to a derivative of a product, \( f(u)g'(u) - g(u)f'(u)\). This is because it relates to the concept of **integration by observation**, where identifying familiar derivative patterns can immensely simplify finding the integral.
Understanding how to identify these structures:

• Allows you to utilize various integration techniques more effectively.
• Enables you to simplify integrals by recognizing underlying patterns and relationships between functions and their derivatives.

This concept was crucial in approaching the original exercise, transforming it into a deceptively simpler form that led directly to the solution.