Question

\(\int[\mathrm{dx} / \sqrt{(x-4)(x-7)}]=+c \quad(4

Short answer

The short answer is: \[\boxed{2 \sin^{-1}\left(\sqrt{\frac{x - 4}{3}}\right) + C}\]

Step-by-step solution

Identify the integral

Given the integral: \[\int\frac{dx}{\sqrt{(x-4)(x-7)}}\]

Perform a substitution

Let \(u = \frac{x - 4}{3}\), which implies \(x = 3u + 4\). Then differentiate x in terms of u: \[\frac{dx}{du} = 3, \quad\Rightarrow dx = 3 du\] Now, substitute the value of x and dx in the integral: \[\int\frac{3 du}{\sqrt{(3u)(3u-3)}}\]

Simplify the integrand

The expression becomes: \[\int\frac{3 du}{\sqrt{9u(3u-3)}} = \int\frac{3 du}{\sqrt{9u^2(3-u)}}\] Now, factor out the 9 and take the square root: \[\int\frac{3 du}{3 \sqrt{u(3-u)}}\] Cancel out the 3: \[\int\frac{du}{\sqrt{u(3-u)}}\]

Apply an inverse trigonometric function identity

The given integral reminds us of the following identity: \[\frac{d}{dx}\sin^{-1}\left(\frac{x}{a}\right) = \frac{1}{\sqrt{a^2 - x^2}}\] Comparing the integrand, we see that: \[a^2 - x^2 = 3 - u \quad\Rightarrow (1/3)(a^2 - u^2) = (1/3)(3 - u)\] Setting \(a^2 = 3\), we have: \[\int\frac{du}{\sqrt{(1/3)(a^2 - u^2)}}\] Which is similar to the identity, so we can use the inverse sine function.

Find the antiderivative

From the identity, we have: \[2 \sin^{-1}\left(\frac{u}{\sqrt{3}}\right) + C\] Now, substitute back for x: \[2 \sin^{-1}\left(\frac{\frac{x - 4}{3}}{\sqrt{3}}\right) + C = 2 \sin^{-1}\left(\sqrt{\frac{x - 4}{3}}\right) + C\] This matches option (a) \[\boxed{2 \sin^{-1}\left(\sqrt{\frac{x - 4}{3}}\right) + C}\]

Key concepts

Substitution Method in Integration

The substitution method in integration is used to simplify complex integrals by transforming them into more manageable forms. In our problem, the algebraic expression \(\sqrt{(x-4)(x-7)}\) presented a challenging integrand. The key idea is to introduce a new variable that will simplify this expression.

We started by setting \(u = \frac{x - 4}{3}\), which simplifies to \(x = 3u + 4\).
This step helps us shift from a complex variable \(x\) to a simpler variable \(u\).
By differentiating, \(\frac{dx}{du} = 3\), we can express \(dx\) in terms of \(du\) as \(dx = 3 du\).

Substitution reduces the original integral into:

• Rewriting \(dx\) and adjusting the limits to the new variable.
• Simplifying the integrand to focus on solving a standard form.

Through substitution, the integral becomes easier to handle and can be solved with basic integration techniques.

Inverse Trigonometric Functions

Inverse trigonometric functions come in handy when dealing with integrals of certain forms. In this problem, we restructured the integrand to resemble a standard format that utilizes \(\sin^{-1}\), also known as the arcsine function.

The function \[\frac{du}{\sqrt{u(3-u)}}\] suggests a comparison with the derivative of the inverse sine function, specifically \[\frac{d}{dx}\sin^{-1}\left(\frac{x}{a}\right) = \frac{1}{\sqrt{a^2 - x^2}}\].
This expression allows us to directly integrate using the identity, thereby simplifying the antidifferentiation process.

In our final expression, we used the inverse sine function, a powerful tool for solving integrals involving quadratic expressions within a square root in the denominator. This connection greatly simplifies the solution and helps to find the antiderivative efficiently.

Integration Techniques

Integration techniques such as substitution and the application of inverse trigonometric functions enable us to solve complex problems effectively.

In this problem, combining substitution with inverse trigonometric identities allowed us to solve an otherwise intricate integral. This systematic approach transforms the integrand into a standard form that is straightforward to integrate using known formulas.

The use of these techniques involves:

• Recognizing suitable substitutions that neutralize complications within the integral.
• Utilizing known identities like those of trigonometric derivatives for simplification.

Through such methods, complicated, seemingly inaccessible integrals become feasible, making these techniques invaluable in calculus.

JEE Mathematics

The Joint Entrance Examination (JEE) Mathematics section often includes problems that test knowledge of advanced concepts like integration. Understanding the use of substitution and inverse trigonometric functions is crucial in solving these problems.

In our exercise, the integral presented challenges typical of JEE level questions, requiring practical application of integration methods and a deep understanding of trigonometric identities.

To excel in JEE Mathematics, students should focus on:

• Mastering core integration techniques and the conditions for their application.
• Solving a wide variety of practice problems to familiarize with common tricks and transformations.
• Developing an intuition for when and how to use specific methods, like substitution or inverse functions, effectively.

Grasping these concepts will significantly improve confidence and efficiency in tackling similar calculus problems on exams.