Problem 428 - Calculus Volume 1

Question

In the following exercises, compute each integral using appropriate substitutions. $$\int \frac{e^{t} \cos ^{-1}\left(e^{t}\right)}{\sqrt{1-e^{2 t}}} d t$$

Step-by-Step Solution

Answer

\(-e^t \cos^{-1}(e^t) + \sqrt{1-e^{2t}} + C\)

Step 1: Identify the Substitution Variable

The integral is of the form \( \int \frac{e^{t} \cos^{-1}(e^{t})}{\sqrt{1-e^{2t}}} \, dt \). Here, note the presence of expressions involving \( e^t \). Let's set \( u = e^{t} \). Therefore, \( du = e^t \, dt \) or equivalently \( dt = \frac{du}{u} \). This substitution simplifies the integrand significantly due to the relationship \( e^t = u \).

Step 2: Substitute into the Integral

Using the substitution \( u = e^t \), the integral becomes \( \int \frac{u \cos^{-1}(u)}{\sqrt{1-u^2}} \cdot \frac{du}{u} \). Simplifying this, the \( u \) terms cancel out, yielding \( \int \frac{\cos^{-1}(u)}{\sqrt{1-u^2}} \, du \).

Step 3: Recognize the Integral Form

Notice that the integral \( \int \frac{\cos^{-1}(u)}{\sqrt{1-u^2}} \, du \) corresponds to a standard integral formula. Specifically, it integrates directly to \(-u \cdot \cos^{-1}(u) + \sqrt{1-u^2} + C\), where \( C \) is the constant of integration.

Step 4: Back Substitute the Original Variable

Since we used the substitution \( u = e^t \), we need to return to the variable \( t \). The result of the integral in terms of \( t \) is then \(-e^t \cdot \cos^{-1}(e^t) + \sqrt{1-e^{2t}} + C\). Therefore, the integral \( \int \frac{e^{t} \cos ^{-1}(e^{t})}{\sqrt{1-e^{2 t}}} d t \) simplifies to this expression.

Key Concepts

Substitution Method

The substitution method is a magical tool in calculus that simplifies solving complex integrals. It's like changing a foreign movie into your native language—everything becomes easier to understand. In integrals, this means finding a new variable to replace part of the original expression, making it easier to solve. Let's break it down step-by-step.

When using the substitution method:

• Identify a part of the integral that complicates the process. This is the section you want to substitute because it will simplify the math.
• Introduce a new variable, say \( u \), which will substitute the identified part. Define it to express the complicated part in simpler terms, like \( u = f(x) \).
• Differentiate your new variable, \( du \), to account for the change in terms of \( dx \). This involves finding the derivative of \( u \) with respect to \( x \). It will allow you to replace \( dx \) in the integral.
• Substitute all parts of the original integral using \( u \) and \( du \), including changing the limits in definite integrals.
• Solve the simplified integral with respect to \( u \). It’s usually a known form or much easier to solve.
• Finally, back-substitute the original variable to express the solution in terms of the initial variable.

Using substitution not only simplifies the problem but also gives a fresh perspective on the original integral's nature.

Trigonometric Integrals

Trigonometric integrals appear daunting, but they are a staple in calculus. They involve integrals of trigonometric functions, which often require specific strategies to solve. Imagine navigating a maze—knowing trigonometric identities can be the map that guides you out.

Here are some strategies:

• Use trigonometric identities to simplify the expression before integration. Functions like \( \sin^2(x) \), \( \cos^2(x) \), and \( \tan(x) \) often convert into other forms using identities such as \( \sin^2(x) + \cos^2(x) = 1 \).
• Sometimes, substitutions involving trigonometric identities can convert a difficult integrand into a simpler one. For instance, substituting \( \sin(x) = u \) might help.
• If you’re dealing with powers of sine and cosine, try reducing the powers using identities like the half-angle formulas.
• Look for recognizable patterns or standard integral forms. Many trig functions integrate directly or via substitution into known results.

Simplifying trigonometric integrals requires these techniques, along with practice, to recognize the integral forms swiftly.

Definite and Indefinite Integrals

Integrals are fundamental in calculus, representing the accumulated sum or area under a curve. They come in two flavors: definite and indefinite, each with unique characteristics and applications.

Indefinite Integrals:

• Represent the family of functions, or antiderivatives, of a given function. It's like knowing all the possible pasts for a character in a story.
• The result includes a constant \( C \), since differentiation of a constant is zero, and thus it's part of all antiderivatives.
• In notation, written as \( \int f(x) \, dx \), meaning the antiderivative of \( f(x) \).

Definite Integrals:

• Calculate the exact area under a curve between two points, denoted as \([a, b]\). It’s the actual accumulated value rather than just possibilities.
• The result has no \( C \) since it's a numeric value.
• The Fundamental Theorem of Calculus links the two: evaluate the antiderivative at the boundaries \( a \) and \( b \) and find the difference.

Both types of integrals are essential in various fields—physics, engineering, economics—for solving real-world problems by capturing change and accumulation.