Problem 44 - Calculus: One and Several Variables

Question

Derive the formula.$$\int \frac{u d u}{a^{2}-u^{2}}=-\frac{1}{2} \ln \left|a^{2}-u^{2}\right|+C$$.

Step-by-Step Solution

Answer

To derive the given formula, perform the substitution \(a^2 - u^2 = t\), differentiate with respect to \(u\) to find \(\frac{dt}{2} = -u\,du\), and rewrite the integral using the substitution. Simplify and evaluate the integral, then substitute back to get the final expression: \(-\frac{1}{2} \ln|a^2 - u^2| + C\).

Step 1: Substitution

To solve this integral, we will perform the substitution \(a^2 - u^2 = t\). So, we have \(t = a^2 - u^2\). To find the derivative of \(t\) with respect to \(u\), we differentiate both sides with respect to \(u\). \[\frac{dt}{du} = -2u\] Then, we have \(\frac{dt}{2} = -u\,du\). Now, let's rewrite the given integral using this substitution. #Step 2: Rewrite the integral#

Step 2: Rewrite the integral

Using the substitution from step 1, we can rewrite the original integral as: \[\int \frac{u\,du}{a^2 - u^2} = \int\frac{-\frac{dt}{2}}{t}\] Now, simplify the integral: \[-\frac{1}{2} \int\frac{dt}{t}\] #Step 3: Evaluate the integral#

Step 3: Evaluate the integral

Integrating with respect to \(t\), we get: \[-\frac{1}{2} \ln|t| + C\] #Step 4: Substitute back#

Step 4: Substitute back

In step 1, we made the substitution \(t = a^2 - u^2\). Now, let's substitute back to find the final solution: \[-\frac{1}{2} \ln|a^2 - u^2| + C\] This is the expected final expression, so we have derived the desired formula successfully.

Key Concepts

U-Substitution

When dealing with integral calculus, 'u-substitution' is a critical skill for simplifying and solving complex integrals. It's a technique that allows us to transform an integral into a simpler form by substituting a portion of the integral with a new variable, commonly denoted as 'u'. The essence of u-substitution is finding an appropriate function and its derivative within the integral that, when replaced with 'u', the integral becomes easy to manage.

For example, if you encounter an integral like \[\int x \cdot \sin(x^2)\,dx\], you may let 'u = x^2'. Then, finding the differential 'du' (which is '2x\,dx' in this case), you can rewrite the integral in terms of 'u', making it simpler to integrate.

Note that after solving for the new variable, 'u', you must substitute back the original variable to complete the solution. This step is crucial for the integrity of the answer.

Indefinite Integrals

Indefinite integrals are integrals that do not specify the limits of integration, representing a family of functions rather than a numeric value. The notation used for an indefinite integral is \(\int f(x)\,dx\), and the solution is generally expressed as a function plus an arbitrary constant of integration, denoted as 'C'. This constant accounts for the unknown quantity that is added during the anti-differentiation process.

The solution to an indefinite integral represents the antiderivative of the function. The importance of finding the antiderivative lies in its ability to describe accumulation and area under a curve, which is fundamental in various fields such as physics, engineering, and economics.

An important note for students is to remember to always include the constant 'C' at the end of every indefinite integral to signify that there are an infinite number of antiderivatives.

Integration Techniques

In the realm of calculus, 'integration techniques' consist of various methods used to solve integrals that are not readily solvable by a simple antiderivatives table or basic rules. Apart from u-substitution, there are several other techniques such as integration by parts, partial fraction decomposition, trigonometric substitution, and more. Each method serves a specific type of integral or integrals containing certain functions.

For instance, 'integration by parts' is useful when dealing with products of functions, while 'partial fractions' can simplify rational functions into simpler fraction components. Understanding when and how to apply each technique is essential, and often, recognizing patterns and strategic manipulation of the integral are key to selecting the appropriate method.

For students, practice is vital. Being able to identify which method to use rapidly will save time and prevent mistakes during exams or in practical applications.

Natural Logarithm Properties

The natural logarithm, denoted as 'ln', has properties that are instrumental in solving calculus problems, especially when dealing with logarithmic differentiation and integration. Some critical properties include:\

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• The logarithm of a product is the sum of the logarithms: \(ln(ab) = ln(a) + ln(b)\).
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• The logarithm of a quotient is the difference of the logarithms: \(ln(\frac{a}{b}) = ln(a) - ln(b)\).
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• The logarithm of a power is the power times the logarithm: \(ln(a^k) = k\cdot ln(a)\).
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• For the natural logarithm, \(ln(e) = 1\) and \(ln(1) = 0\).
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Understanding these properties enables students to handle integrals involving natural logarithms, such as solving \[\int \frac{1}{x} \,dx\], which results in \(ln|x| + C\). They also provide the groundwork for logarithmic differentiation, which is a method used for differentiating functions in the form of one variable raised to the power of another.