Problem 67 - University Calculus: Early Transcendentals

Question

If you do not know what substitution to make, try reducing the integral step by step, using a trial substitution to simplify the integral a bit and then another to simplify it some more. You will see what we mean if you try the sequences of substitutions. \(\int \frac{18 \tan ^{2} x \sec ^{2} x}{\left(2+\tan ^{3} x\right)^{2}} d x\) a. \(u=\tan x,\) followed by \(v=u^{3},\) then by \(w=2+v\) b. \(u=\tan ^{3} x,\) followed by \(v=2+u\) c. \(u=2+\tan ^{3} x\)

Step-by-Step Solution

Answer

All methods result in \(-\frac{6}{2+\tan^3 x} + C\).

Step 1a: Using Substitution 1 in option (a)

First, we use the substitution \( u = \tan x \). The derivative \( \frac{d}{dx} (\tan x) = \sec^2 x \), so \( du = \sec^2 x \, dx \). The integral becomes:\[\int \frac{18 u^2 \sec^2 x}{(2 + u^3)^2} \sec^2 x \, dx = \int \frac{18 u^2}{(2 + u^3)^2} \, du\]

Step 1b: Using Substitution 2 in option (a)

Next, apply the substitution \( v = u^3 \), then \( dv = 3u^2 \, du \). Thus, \( du = \frac{dv}{3u^2} \). Substituting we have:\[\int \frac{18 u^2}{(2 + v)^2} \cdot \frac{1}{3u^2} \, dv = \int \frac{6}{(2+v)^2} \, dv\]

Step 1c: Using Substitution 3 in option (a)

Finally, substitute \( w = 2 + v \), then \( dw = dv \). The integral simplifies to:\[\int \frac{6}{w^2} \, dw\]This integral can be solved directly, resulting in:\[-\frac{6}{w} + C = -\frac{6}{2+u^3} + C = -\frac{6}{2+\tan^3 x} + C\]

Step 4: Conclusion for Option (a)

After a sequence of substitutions, we obtained the integral as:\[-\frac{6}{2+\tan^3 x} + C\] which is the final simplified form of the integral using option (a).

Step 2: Using Substitution in option (b)

Use the substitution \( u = \tan^3 x \), so \( du = 3 \tan^2 x \sec^2 x \, dx \). Rearrange to find \( dx \):\[ dx = \frac{du}{3 \tan^2 x \sec^2 x} \]Substitute into the integral:\[\int \frac{18 \tan^2 x \sec^2 x}{(2+u)^2} \cdot \frac{du}{3 \tan^2 x \sec^2 x} = \int \frac{6}{(2+u)^2} \, du\]Solving this gives:\[-\frac{6}{2+u} + C = -\frac{6}{2+\tan^3 x} + C\]

Step 6: Conclusion for Option (b)

Using option (b) also simplifies the integral to:\[-\frac{6}{2+\tan^3 x} + C\]

Step 3: Using Substitution in option (c)

With \( u = 2 + \tan^3 x \), we have \( du = 3 \tan^2 x \sec^2 x \, dx \). Rearrange to find \( dx \):\[ dx = \frac{du}{3 \tan^2 x \sec^2 x} \]Substitute in the given integral:\[\int \frac{18 \tan^2 x \sec^2 x}{u^2} \cdot \frac{du}{3 \tan^2 x \sec^2 x} = \int \frac{6}{u^2} \, du\]The solution to this is:\[-\frac{6}{u} + C = -\frac{6}{2+\tan^3 x} + C\]

Step 8: Conclusion for Option (c)

Option (c) also leads us to:\[-\frac{6}{2+\tan^3 x} + C\], the same solution as for options (a) and (b).

Key Concepts

Integral Calculus

Integral calculus is a fundamental part of mathematics that focuses on the concept of integration. Integration is used to find areas under curves, accumulated quantities, and functions that describe the total change from a derivative. In integral calculus, we deal with two main operations: definite integrals and indefinite integrals. A definite integral gives the exact area under a curve between two points while an indefinite integral is more abstract, representing a family of functions. The notation for an indefinite integral is written as \( \int f(x) \, dx \), where \( f(x) \) is the integrand and \( dx \) represents the variable of integration.

Integration often involves reversing the process of differentiation and finding a function given its derivative. This reverse process is what makes integral calculus a vital tool in analyzing and understanding changes in systems. Remember that every time you solve an integral, adding a constant \( C \) is essential because there are infinitely many antiderivatives for any given function.

Trigonometric Integrals

Trigonometric integrals involve integrating functions that include trigonometric functions such as sine, cosine, tangent, or their powers. These integrals often require special techniques for simplification due to the periodic and oscillatory nature of trigonometric functions.

Some common methods to handle trigonometric integrals include using identities, like \( \sin^2(x) + \cos^2(x) = 1 \), as well as substitution strategies that convert the integral into a simpler form. For instance, when dealing with an integral involving \( \tan(x) \) and \( \sec(x) \), like the one in our exercise, substituting \( \tan(x) \) or \( \sec(x) \) to simplify the expression may be effective. These methods reduce complex expressions to more manageable forms, making the integration process much simpler.

Understanding the underlying identities and properties of trigonometric functions is crucial in solving these integrals efficiently. Practice different approaches to gain fluency, as being versatile with these concepts will be helpful across various calculus problems.

Change of Variables

The change of variables, or substitution, is a powerful technique in calculus that simplifies complicated integrals by transforming them into more familiar or manageable forms. This technique involves replacing a variable in the integral with a new variable that represents the same quantity but is expressed differently. By doing this, we highlight the structure of the integral or expose a pattern that can be easily integrated.

In the given textbook solution, several substitutions were used consecutively to simplify the integral into a straightforward form. Each step in the substitution process aims to progressively ease the complexity of the integration. Here’s how it usually works:

• Choose a substitution that simplifies the integrand, typically by making a part of it a direct derivative.
• Rewrite the integral in terms of the new variable, changing all occurrences of the original variable, including the \( dx \).
• Solve the new, simpler integral with respect to the new variable, then substitute back the original variable at the end if necessary.

This technique is very versatile and can often transform an otherwise difficult problem into one with a known solution. Leveraging substitution effectively requires recognizing patterns and understanding how changes in variable affect the integral's form.