Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use the table of integrals on the inside back cover, perhaps combined with a substitution, to evaluate the given integrals. $$ \int \frac{\cos t \sin t}{\sqrt{2 \cos t+1}} d t $$

Step-by-Step Solution

Answer

The integral evaluates to \( -\frac{1}{6}(2\cos t + 1)^{3/2} + \frac{1}{2}(2\cos t + 1)^{1/2} + C \).

Step 1: Identify the Substitution

Notice that the integrand contains a composite function involving \( \cos t \). Therefore, let's choose a substitution to simplify the integral. Set \( u = 2\cos t + 1 \). Then, differentiate \( u \) to find \( du \).

Step 2: Calculate \( du \) in terms of \( dt \)

Given \( u = 2\cos t + 1 \), differentiate with respect to \( t \): \( \frac{du}{dt} = -2\sin t \). This implies \( du = -2\sin t \,dt \), or equivalently, \( -\frac{1}{2} du = \sin t \, dt \).

Step 3: Rewrite the Integral

Substitute \( u = 2\cos t + 1 \) and \( \sin t \, dt = -\frac{1}{2} \, du \) into the integral. The integral becomes \( \int \frac{\cos t}{\sqrt{u}} \cdot -\frac{1}{2} \, du \).

Step 4: Express \( \cos t \) in terms of \( u \)

Since \( u = 2\cos t + 1 \), solve for \( \cos t \): \( \cos t = \frac{u - 1}{2} \). Substitute this into the integral to get \( -\frac{1}{2} \int \frac{\frac{u - 1}{2}}{\sqrt{u}} \, du \).

Step 5: Simplify the Integral

Simplify the expression to \( -\frac{1}{4} \int \frac{u - 1}{\sqrt{u}} \, du \). This can be further broken down as \( -\frac{1}{4} \left( \int \sqrt{u} \, du - \int \frac{1}{\sqrt{u}} \, du \right) \).

Step 6: Integrate Each Term Separately

The integrals to calculate are \( \int u^{1/2} \, du = \frac{2}{3}u^{3/2} \) and \( \int u^{-1/2} \, du = 2u^{1/2} \). Substitute these results back into the expression for the original integral.

Step 7: Combine and Simplify

Substitute back: \( -\frac{1}{4} \left( \frac{2}{3}u^{3/2} - 2u^{1/2} \right) = -\frac{1}{6}u^{3/2} + \frac{1}{2}u^{1/2} \).

Step 8: Substitute \( u \) back in Terms of \( t \)

Remember that \( u = 2\cos t + 1 \). Substitute back into the expression: \( -\frac{1}{6}(2\cos t + 1)^{3/2} + \frac{1}{2}(2\cos t + 1)^{1/2} + C \), where \( C \) is the constant of integration.

Key Concepts

Substitution Method

The substitution method is a powerful technique used in calculus to simplify integrals, especially when dealing with composite functions. If you notice a function inside another function within your integral, like in our exercise, this method might be your best friend. The main idea is to replace the composite part of the function with a single variable.

Here's how it works:

• Choose a substitution. Find a part of your integral that can be substituted to make the integration process smoother. For example, if you have 2cos(t) + 1, you might let u = 2cos(t) + 1.
• Differential replacement. Calculate the differential of your substitution, in this case, du = -2sin(t) dt. Adjust your original integral using this new differential relationship.
• Replace and simplify the integral. Switch out the corresponding parts in your original integral with your substitution and new differential expression.

This transforms a complex integral into something more manageable by reducing its components, often into a standard form that's easier to handle or even listed in an integral table.

Integral Tables

Integral tables are a collection of standard integrals which help students and professionals find solutions without having to integrate from scratch. These tables are essential for tackling a variety of integrals, particularly when they appear in complicated combinations. When substitution is applied, it can transform an unusual integral into a format found within these tables.

Using integral tables effectively:

• Identify the form. After using methods like substitution, compare the modified integral with the forms listed in your table to see if it matches any standard integral.
• Follow the rules. Some integrals in the table may require additional algebraic manipulation to match the form found in your problem.
• Shortcut those tough calculations. Recognizing a standard form in your table reduces the need for calculative rigor, allowing you to skip certain steps and get immediate solutions.

This table-based approach is incredibly useful when faced with complex integrals, saving time and effort by utilizing pre-established solutions.

Trigonometric Integration

Trigonometric integration is another technique applied to integrals involving trigonometric functions like sin, cos, tan, etc. Given the integral we looked at, the involved functions were sin(t) and cos(t), which are typical characters in trigonometric expressions.

Here's what to keep in mind:

• Recognize common identities. Use trigonometric identities to simplify the equation. For instance, knowing that sin^2(x) + cos^2(x) = 1 can sometimes simplify or transform an integral.
• Utilize substitutions. As seen in the exercise, introducing a substitution (like u for 2cos(t) + 1) might turn a trigonometric problem into an algebraic one.
• Pattern spotting. Be on the lookout for patterns that match known integral forms involving trigonometric functions.

Trigonometric integration requires practice to become familiar with these simplifications and transformations, making handling angles and periodic functions more intuitive.

Composite Functions in Integration

Composite functions, a function within another function, often appear in integration problems. In calculus, decomposing or transforming these functions is essential for straightforward integration.

Here's how composite functions were managed in this exercise:

• Detection. Spotting composite functions like 2cos(t) + 1 helps identify the opportunity to apply the substitution method, simplifying the integral.
• Find expressions for replacements. Determine expressions in terms of your new substitution variable (u) rather than the original variable (t).
• Simplification. Using the substitution method not only simplifies the integral but can also change a trigonometric problem into simpler algebraic terms.

Handling composite functions effectively opens avenues to both identifying simpler forms of expression and understanding the deeper structure of integrals for quicker integration solutions.