Problem 12 - Applied Calculus

Question

Find each integral. $$ \int \cos (1-t) d t $$

Step-by-Step Solution

Answer

The integral of \( \cos (1-t) \) with respect to \( t \) is \(-\sin(1-t) + C\).

Step 1: Identify the Integral Type

The given integral \( \int \cos (1-t) \, dt \) involves a trigonometric function of a linear expression. It fits the pattern \( \int \cos (u) \, du \), where \( u = 1-t \). To solve it, we will use substitution.

Step 2: Substitution Setup

Set \( u = 1-t \). Then, differentiate both sides with respect to \( t \) to find the differential: \( du = -dt \), or equivalently, \( dt = -du \). This will help us rewrite the integral in terms of \( u \).

Step 3: Substitute and Simplify

Substitute \( u = 1-t \) and \( dt = -du \) into the integral to obtain:\[\int \cos(u) \, (-du) = - \int \cos(u) \, du\]The negative sign comes from the substitution \( dt = -du \).

Step 4: Integrate with respect to u

The integral of \( \cos(u) \) with respect to \( u \) is \( \sin(u) + C \), where \( C \) is the constant of integration. Therefore, we have:\[- \int \cos(u) \, du = - (\sin(u) + C)\]Which simplifies to:\[-\sin(u) + C\]

Step 5: Back-Substitute to Original Variable

Replace \( u \) with \( 1-t \) to express the integral back in terms of \( t \):\[-\sin(1-t) + C\]This is the indefinite integral in terms of \( t \).

Key Concepts

Substitution Method

The substitution method is a powerful tool in calculus that simplifies complex integrals. It involves changing the variable of integration to transform the integral into an easier form. Essentially, it's about finding a substitution that simplifies the integrand, often when a chain rule appears in reverse.

In our example, we identify that the cosine function's argument is a linear expression, which hints at substitution as a viable strategy. We set \( u = 1-t \), which simplifies our integrand. After differentiation, we find \( du = -dt \), or equivalently \( dt = -du \). By substituting \( u \) and \( dt \) in the integral, we convert \( \int \cos(1-t) \, dt \) into \(- \int \cos(u) \, du \).

Key steps in the substitution method include:

• Identifying a suitable substitution where a part of the integrand can be replaced by a new variable.
• Differentiating to express the original differential in terms of the new differential.
• Rewriting the entire integral in terms of the new variable, making it easier to integrate.

By performing these steps, you can transform complex integrals into simpler ones that are more straightforward to solve.

Indefinite Integrals

Indefinite integrals, sometimes known as antiderivatives, represent a family of functions whose derivative is the integrand. Unlike definite integrals, which compute area under a curve between specific limits, indefinite integrals do not have limits of integration.

Indefinite integrals are denoted by \( \int f(x) \, dx \), where \( f(x) \) is the integrand, and \( dx \) signifies integration with respect to the variable \( x \). The result of an indefinite integral includes a constant \( C \), as integration represents an accumulation function, and this constant accounts for all possible vertical shifts of the antiderivative.

In our example of \( \int \cos(u) \, du \), the indefinite integral is \( \sin(u) + C \), reflecting the general rule that the antiderivative of \( \cos(u) \) is \( \sin(u) \). When we solve indefinite integrals, we provide a general solution inclusive of the constant \( C \), which signifies any possible starting point of the integrated function.

Understanding indefinite integrals is essential in calculus for solving differential equations, finding solution families, and analyzing functions from their rate of change.

Trigonometric Functions

Trigonometric functions are fundamental in mathematics, especially in calculus, because they model periodic phenomena such as sound and light waves. In integration, knowing the integral forms of trigonometric functions is crucial for evaluating expressions involving them.

The primary trigonometric functions include sine (\( \sin \)), cosine (\( \cos \)), and tangent (\( \tan \)). For integration purposes, it is essential to remember their derivatives and integrals:

• The derivative of \( \sin(x) \) is \( \cos(x) \), and its integral is \(-\cos(x) + C\).
• The derivative of \( \cos(x) \) is \(-\sin(x)\), and its integral is \( \sin(x) + C\).

In our exercise, the integrand \( \cos(1-t) \) involves cosine, so we rely on the fact that its antiderivative is sine. Furthermore, the result of this integral, after applying substitution and considering the negative sign from \( dt = -du \), yields \(-\sin(1-t) + C\).

Mastering these basic integral forms is an integral part of learning calculus, simplifying many advanced mathematical problems.