Problem 50 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(39-56\) $$ \int \frac{\sec y \tan y}{2+\sec y} d y $$

Step-by-Step Solution

Answer

\( \ln|2 + \sec y| + C \)

Step 1: Rewrite the Integral

Given the integral \( \int \frac{\sec y \tan y}{2 + \sec y} dy \), the first step is to identify if there are any substitutions or simplifications that can be made to simplify the problem. In this problem, consider a substitution that can simplify the denominator.

Step 2: Choose a Suitable Substitution

We notice that the denominator is of the form \(2 + \sec y\). A clever substitution here is \(u = 2 + \sec y\), which would make \(\frac{du}{dy} = \sec y \tan y\). This simplifies the integrals considerably.

Step 3: Differentiate to Find \(du\)

Differentiate the substitution \(u = 2 + \sec y\). This gives us \(\frac{du}{dy} = \sec y \tan y\), which implies that \(du = \sec y \tan y \, dy\). This means \(dy = \frac{du}{\sec y \tan y}\).

Step 4: Substitute and Simplify the Integral

Substitute the expressions found in the previous steps into the integral. Thus, the integral becomes \( \int \frac{\sec y \tan y}{u} \cdot \frac{du}{\sec y \tan y} = \int \frac{1}{u} du \). The \(\sec y \tan y\) terms cancel each other, leaving a simple integrand.

Step 5: Evaluate the Integral

The simplified integral is \(\int \frac{1}{u} du\), which is a basic integral that evaluates to \(\ln|u| + C\).

Step 6: Substitute Back and Finalize the Solution

Substitute back the original variable: \(u = 2 + \sec y\). Thus, the solution becomes \(\ln|2 + \sec y| + C\), where \(C\) is the constant of integration.

Key Concepts

Substitution Method

The substitution method is a pivotal technique in calculus, especially for evaluating integrals. It involves replacing a complicated expression with a single variable, usually denoted as \( u \), to make calculations simpler.

• Identify an expression within the integral that can be replaced with \( u \). This expression is often part of the integrand that repeats or complicates the integral.
• Determine the corresponding derivative \( du \) in terms of \( dy \) (or the original variable) to properly substitute the differential part of the integrand.
• Substitute these expressions back into the integral, transforming it into a simpler form.

For our example, we chose \( u = 2 + \sec y \) because it simplifies the integrand considerably. This substitution neutralizes the complex trigonometric term in the denominator. The transformation into \( \int \frac{1}{u} \, du \) allows us to integrate easily, showing the power of substitution.

Integrals

An integral represents the area under a curve or the accumulation of quantities over a continuous interval. There are different techniques to solve integrals, each suitable for different forms.

• Indefinite integrals, such as \( \int f(y) dy \), provide a family of functions whose derivative is \( f(y) \).
• Definite integrals, \( \int_{a}^{b} f(y) dy \), compute the area between the curve \( f(y) \), the x-axis, and the vertical lines \( y = a \) and \( y = b \).

In this exercise, we dealt with an indefinite integral, transforming \( \int \frac{\sec y \tan y}{2 + \sec y} dy \) into \( \int \frac{1}{u} du \). Integration of \( \frac{1}{u} \) results in \( \ln|u| + C \), a classic example of the natural logarithm's involvement in integration.

Trigonometric Identities

Trigonometric identities allow us to express trigonometric functions in various useful forms. These identities are essential when tackling integrals involving trigonometric expressions.

• Common identities include \( \sin^2 y + \cos^2 y = 1 \) and \( \sec y = \frac{1}{\cos y} \).
• They enable the simplification of complex trigonometric integrands into more manageable algebraic expressions.

In the given example, the trigonometric function \( \sec y \tan y \) in the integrand is critical to the substitution. Our specific identity \( \frac{d}{dy}(\sec y) = \sec y \tan y \) directly facilitated the substitution process, illustrating how identities are used practically to solve integrals.

Constant of Integration

When we evaluate indefinite integrals, we add a constant, \( C \), to express all possible antiderivatives. Since differentiation of a constant results in zero, any antiderivative \( F(y) + C \) differentiates to the original function \( f(y) \).

• The constant of integration, \( C \), accounts for shifts in the function vertically on a graph.
• Mathematically, it represents any potential vertical displacement of the primitive function.

For the integral \( \int \frac{1}{u} du \), solving yields \( \ln|u| + C \). Upon reverting from \( u \) back to the original variable, the solution becomes \( \ln|2 + \sec y| + C \), emphasizing that though \( u \) changes back to \( y \)-terms, \( C \) remains as the integration constant.