Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 13-30, use the table of integrals on the inside back cover, perhaps combined with a substitution, to evaluate the given integrals. (a) \(\int \frac{d z}{z \sqrt{5-4 z}}\) (b) \(\int \frac{\sin x}{\cos x \sqrt{5-4 \cos x}} d x\)

Step-by-Step Solution

Answer

(a) \(-\frac{1}{5-4z}\sqrt{5-4z} + C\); (b) \(-\frac{1}{5-4\cos x}\sqrt{5-4\cos x} + C\)."

Step 1: Identify the table of integrals

Look for a formula in the table of integrals that is applicable to integrals of the form \( \int \frac{1}{u \sqrt{a - bu}} \, du \). If such a formula exists, it will simplify the process of evaluating these integrals.

Step 2: Substitution for Integral (a)

For the integral \( \int \frac{d z}{z \sqrt{5-4z}} \), use the substitution \( u = 5 - 4z \), which means \( du = -4 \, dz \) or \( dz = -\frac{1}{4} du \). Therefore, the limits of integration change, and the integral can be rewritten in terms of \( u \).

Step 3: Rewrite Integral (a) in Terms of u

Substitute the expressions from Step 2 into the integral: \( \int \frac{d z}{z \sqrt{u}} = -\frac{1}{4} \int \frac{1}{\frac{(5-u)}{4}\sqrt{u}} \, du \), which simplifies to \(-\frac{1}{4} \int \frac{4}{5-u} \cdot \frac{1}{\sqrt{u}} \, du \).

Step 4: Simplify and Solve Integral (a)

The integral is similar to \( -\int \frac{du}{u \sqrt{u}} \), which can be solved by recognizing it fits a standard integral. Applying the formula from the table of integrals gives \( -\frac{1}{5-u} \sqrt{5-4z} \) plus the constant of integration, \( C \).

Step 5: Substitution for Integral (b)

For the integral \( \int \frac{\sin x}{\cos x \sqrt{5-4\cos x}} \, dx \), use the substitution \( u = \cos x \), giving \( du = -\sin x \cdot dx \) or \( dx = -\frac{du}{\sin x} \). This makes the integral \( \int \frac{-du}{u \sqrt{5-4u}} \).

Step 6: Simplify and Solve Integral (b)

Simplify the rewritten integral, which is \( \int \frac{du}{u \sqrt{5-4u}} \), matching the form of a standard integral. Using the same table formula, solve to find \( -\frac{1}{5-u} \sqrt{5-4\cos x} \) plus a constant of integration, \( C \).

Step 7: Validate and Combine Results

After solving both integrals using the table of integrals, both results should be re-expressed in terms of their original variables to ensure continuity with the initial problem phrasing.

Key Concepts

Substitution Method

The substitution method is a technique used in integral calculus to simplify the process of finding integrals. It involves substituting a part of the integral with a new variable, making the expression easier to handle. This transformation allows us to recognize and apply standard integrals from a table of known integral solutions.
Here's how you can use substitution:

• Identify a section of your integral that matches a standard form and assign it a new variable, like 'u'. For example, if you have something like \(\sqrt{a - bu}\), you can attempt substituting \(u = g(x)\;\).
• Calculate the 'du' in terms of 'dx' by differentiating your substitution equation. This helps rearrange the differential element \(du = f(x) \, dx\;\).
• Rewrite the original integral in terms of 'u' using the substitution and the new expression for 'dx'. The goal is to make the integral resemble a standard form, facilitating its evaluation with known integrals.

These steps drastically simplify the process of solving complex integrals by reducing them to more manageable forms.

Integral Calculus

Integral calculus is a branch of calculus concerned with finding the accumulation of quantities, which could be areas under curves, volumes, and other meaningful accumulations. Integration is the fundamental concept behind integrating functions.
At its core, integration can be understood as the reverse process of differentiation. Given a function \( f(x)\), its integral \( \int f(x) \, dx\) refers to the collection of all antiderivatives of the function. This process connects directly to finding quantities such as area, volume, and the general accumulation rate of change.
Integration involves several techniques including:

• Substitution, which is often initiated when a direct antiderivative is hard to spot.
• Integration by parts, suitable for products of functions.
• Many more, including partial fraction decomposition and trigonometric substitution.

The purpose of these various techniques is to transform complex integrals into forms that we can match with standard solutions.

Standard Integrals

Standard integrals refer to a list of well-known integral solutions that serve as references for solving calculus problems. These come in handy especially when dealing with complex integrals, allowing us to quickly evaluate integrals that match these standard forms.
You might encounter standard integrals such as:

• The integral of \(\frac{1}{x}\), which is \(\ln |x| + C\,\).
• The integral of \( e^x,\) which remains \( e^x + C\,\).
• There are also more complex standard integrals including trigonometric and exponential forms.

These integrals provide shortcuts by applying direct results from memory or a table without re-deriving them each time. It's crucial to become familiar with these to enhance efficiency in solving calculus problems. Utilizing a table of standard integrals, as seen in the original exercise, allows one to cross-reference and confidently employ these inputs for seamless integration.