Problem 72 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int \frac{d y}{y^{2}+6 y+10}$$

Step-by-Step Solution

Answer

The integral evaluates to \( \arctan(y + 3) + C \).

Step 1: Identify the Integral Type

This integral is of the rational kind in the form \( \int \frac{d y}{a y^2 + b y + c} \). First, we identify that the quadratic expression in the denominator is \( y^2 + 6y + 10 \).

Step 2: Complete the Square

We complete the square for the quadratic expression in the denominator. The expression \( y^2 + 6y + 10 \) can be rewritten by completing the square: \[ y^2 + 6y + 10 = (y + 3)^2 + 1 \] This turns it into \((y + 3)^2 + 1\).

Step 3: Use Trigonometric Substitution

Recognize that the form \( \int \frac{1}{(y + 3)^2 + 1} \, dy \) suggests a trigonometric substitution based on an arctan identity. Let \( u = y + 3 \), which simplifies the integral to \( \int \frac{d u}{u^2 + 1} \).

Step 4: Apply Integral Formula

The integral \( \int \frac{1}{u^2 + 1} \, du \) is a standard form that equals \( \arctan(u) + C \), where \( C \) is the constant of integration. Therefore, \[ \int \frac{d u}{u^2 + 1} = \arctan(u) + C \] Since \( u = y + 3 \), substitute back to get \( \arctan(y + 3) + C \).

Step 5: Final Solution

By substituting back our substitution variable, the integral evaluates to \[ \int \frac{d y}{y^2 + 6y + 10} = \arctan(y + 3) + C \] where \( C \) is the integration constant.

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique used in calculus to evaluate integrals involving certain algebraic expressions, particularly those that contain square roots. In this exercise, however, we applied trigonometric substitution to a rational function by recognizing the pattern in the integral related to the arctan function.
Our integral \[ \int \frac{d y}{y^2 + 6y + 10} \]
first required simplifying its denominator into a recognizable form for trigonometric substitution. Through completing the square, as discussed below, we saw it transform into \((y + 3)^2 + 1\), resembling the form \(a^2 + x^2\), which is suitable for an arctangent identity. Here, the connection brings us to consider the trigonometric identity for arctan, \[ \int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right) + C \]This transformation allows us logically to make a substitution, simplifying the integral so that it fits this standard form, making it easier to solve. By substituting \( u = y + 3 \), our integral becomes easily manageable and aligns with what's known of trigonometric identities.

Completing the Square

Completing the square is a crucial algebraic technique used to rearrange quadratic expressions into a perfect square form. This restructuring helps simplify integration problems, particularly rational functions with quadratic expressions.

• Studying the quadratic expression \(y^2 + 6y + 10\), we aim to express it in the form of \((y + d)^2 + k\).
• The procedure involves taking half of the linear coefficient \(6\), squaring it to get \(9\), and rewriting the expression accordingly.
• Thus, we obtain \((y+3)^2\) which sums to \((y + 3)^2 + 1\) because \(9 + 1 = 10\).

This manipulation simplifies analyzing and performing substitutions so the integral fits trigonometric patterns. It reduces a complex quadratic into something much easier to discuss and integrate, highlighting the importance of this algebraic technique in calculus.

Arctan Identity

The arctan identity is an integral formula that commonly appears in calculus, especially when dealing with the integration of rational functions involving quadratic expressions. The integral
\[ \int \frac{1}{u^2 + 1} \, du \]
is remarkably connected to the arctangent function, resulting directly in \(\arctan(u) + C\) where \(C\) is the integration constant.This insight into arctan identities comes into play when you express a denominator of the form \(x^2 + 1\) or any other such square plus constant form that mimics \((a)^2 + x^2\). By identifying this form after completing the square, you leverage the established integral rules for arctan to swiftly solve the problem. When using this identity, you need to reverse substitute any prior changes made, such as \(u = y + 3\) here, to ensure the solution applies correctly to the original variable of integration, thus maintaining continuity in mathematical representation.