Problem 19 - Thomas Calculus

Question

Evaluate each integral in Exercises \(1-36\) by using a substitution to reduce it to standard form. $$ \int e^{\tan v} \sec ^{2} v d v $$

Step-by-Step Solution

Answer

The integral is \( e^{\tan v} + C \).

Step 1: Identify the substitution

We notice that the integrand contains the expression \( e^{\tan v} \) and \( \sec^2 v \). A useful substitution would target simplifying \( \tan v \), as its derivative is closely related to \( \sec^2 v \). So, let's substitute \( u = \tan v \), which implies \( \frac{du}{dv} = \sec^2 v \).

Step 2: Compute the differential

Based on our substitution \( u = \tan v \), we can express the differential \( dv \) in terms of \( du \). Since \( \frac{du}{dv} = \sec^2 v \), it follows that \( du = \sec^2 v \, dv \).

Step 3: Substitute into the integral

We substitute \( u = \tan v \) and \( du = \sec^2 v \, dv \) into the integral \( \int e^{\tan v} \sec^2 v \, dv \). This gives us \( \int e^u \, du \).

Step 4: Integrate the new integral

The integral \( \int e^u \, du \) is a standard form, which integrates to \( e^u + C \), where \( C \) is the constant of integration.

Step 5: Back-substitute to the original variable

Finally, substitute back the original variable using \( u = \tan v \). Therefore, the antiderivative becomes \( e^{\tan v} + C \).

Key Concepts

Substitution Method

The substitution method is one of the powerful integration techniques used to simplify integrals. It works by transforming the original integral into a simpler one, making the process much more manageable. When faced with complicated expressions inside an integral, such as in the problem \( \int e^{\tan v} \sec ^{2} v d v \), we look for a substitution that can turn the integral into a more recognizable form.
A typical approach is to substitute a part of the integrand with a new variable, often denoted by \( u \). This technique is particularly useful when the derivative of the chosen substitution is present in the integrand, as it allows the replacement of the entire differential.

• Step 1 involves identifying a substitution. Here, because \( \tan v\) is inside the exponential and its derivative, \( \sec^2 v \), is a factor of the integrand, we set \( u = \tan v \).
• Step 2 is calculating the differential. For \( u = \tan v \), the differential is \( du = \sec^2 v \, dv \). This step ensures we can replace \( dv \) with \( du \) in the integral.
• In Step 3, we substitute \( u \) and \( du \) into the integral, converting it into \( \int e^u \, du \).

Indefinite Integrals

Indefinite integrals, also known as antiderivatives, are a type of integral without specified boundaries. They represent a family of functions whose derivative is the integrand. In the context of our problem, after applying the substitution method, we derive the new integral \( \int e^u \, du \), which is an indefinite integral.
For indefinite integrals, the result includes a constant \( C \), symbolizing an infinite set of possible solutions. This constant accounts for the fact that differentiating a constant returns zero, meaning any constant added to or subtracted from a function yields the same derivative.
The process of integrating standard functions like \( e^u \) is straightforward. For instance, \( \int e^u \, du \) evaluates to \( e^u + C \). It's crucial to always return to the original variable after integration in problems involving substitution to provide the final answer in the correct context. Therefore, we substitute back, replacing \( u \) with \( \tan v \), resulting in \( e^{\tan v} + C \).
This expression is the indefinite integral of the original problem, encompassing the entire family of functions that differentiate back to the integrand.

Standard Integrals

Standard integrals are fundamental integral forms that every calculus student should know by heart. Understanding these is essential to quickly solve integrals without lengthy computation. Many integration problems rely on transforming the given integral into one of these standard forms.
In this exercise, after employing the substitution \( u = \tan v \), the integral simplifies into a standard form: \( \int e^u \, du \).
Recognizing standard integrals is important because it allows you to instantly know the antiderivative. Some common standard integrals include:

• \( \int e^x \, dx = e^x + C \)
• \( \int \sin x \, dx = -\cos x + C \)
• \( \int \cos x \, dx = \sin x + C \)

In our discussion, the integral \( \int e^u \, du \) fits the pattern \( \int e^x \, dx = e^x + C \). Hence, we quickly determine its antiderivative as \( e^u + C \).
Harnessing this knowledge simplifies problems significantly, saving time and reducing potential errors.