Problem 75 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(71-84\) $$ \int \frac{d y}{y^{2}-2 y+5} $$

Step-by-Step Solution

Answer

\( \frac{1}{2} \arctan\left(\frac{y-1}{2}\right) + C \)

Step 1: Complete the Square

The denominator of the integral \( y^2 - 2y + 5 \) can be rewritten by completing the square. Begin by factoring the quadratic into the form \((y - 1)^2 + 4\). We find that:\[y^2 - 2y + 5 = (y-1)^2 + 4\]This form will be useful for substitution in later steps.

Step 2: Substitute Variables

To simplify the integral, use the substitution \( u = y - 1 \), which implies \( du = dy \) and the integral becomes:\[\int \frac{du}{u^2 + 4}\]This new integral is in a standard form to use trigonometric identities.

Step 3: Use Trigonometric Substitution

For the integral \( \int \frac{du}{u^2 + 4} \), use \( u = 2 \tan \theta \), where \( du = 2 \sec^2 \theta \, d\theta \). Substitute these into the integral:\[\int \frac{2 \sec^2 \theta \, d\theta}{(2 \tan \theta)^2 + 4} = \int \frac{2 \sec^2 \theta \, d\theta}{4 \tan^2 \theta + 4}\]Simplify to:\[\int \frac{2 \sec^2 \theta \, d\theta}{4(\tan^2 \theta + 1)} = \int \frac{2 \sec^2 \theta \, d\theta}{4 \sec^2 \theta} = \int \frac{1}{2} \, d\theta\]Reduce the integral to a simpler formula.

Step 4: Integrate

Perform the integration:\[\int \frac{1}{2} \, d\theta = \frac{1}{2} \theta + C\]where \( C \) is the constant of integration. This is the integral result in terms of \( \theta \).

Step 5: Back-Substitute to Original Variable

Return to the original variable \( y \). Recall that \( u = 2 \tan \theta \) and \( u = y - 1 \), so:\[y - 1 = 2 \tan \theta \quad \Rightarrow \quad \tan \theta = \frac{y-1}{2}\]Therefore, \( \theta = \arctan\left(\frac{y-1}{2}\right) \). Substitute back to get the final expression:\[\int \frac{dy}{y^2 - 2y + 5} = \frac{1}{2} \arctan\left(\frac{y-1}{2}\right) + C\]

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique used to simplify integrals involving square roots and quadratic expressions. When dealing with integrals of the form \( \int \frac{du}{u^2 + a^2} \), a common approach is to use the substitution \( u = a \tan \theta \). This is beneficial because the identity \( \tan^2 \theta + 1 = \sec^2 \theta \) allows the expression to be rewritten without any squared terms, simplifying the integration process.
This method is particularly useful when the integral resembles a known trigonometric form, enabling the computation of otherwise complex integrals with ease. In our exercise, \( u = 2 \tan \theta \) was used after simplifying the integral for substituting the quadratic expression \( u^2 + 4 \) with a more manageable trigonometric form. This substitution reduced the integral to a simple linear form, illustrating the power of trigonometric identities in calculus.

Completing the Square

Completing the square is a mathematical process used to simplify quadratic expressions. By rewriting the quadratic in the form \( (y - h)^2 + k \), it becomes easier to identify patterns or apply further transformations, such as substitution. For instance, in the original expression \( y^2 - 2y + 5 \), completing the square leads to \( (y - 1)^2 + 4 \). This transformation is often a necessary step before applying trigonometric substitution, as it reveals the structure required for substitution.
The main goal of completing the square is to express the quadratic as a perfect square plus a constant, facilitating operations involving integration and algebraic manipulation. The process involves identifying the linear coefficient, adjusting the constant term, and rewriting the quadratic using perfect-square binomials.

Definite Integration

Definite integration calculates the accumulation of a quantity, often resulting in the area under a curve between two bounds. Although the provided exercise focuses on indefinite integration, understanding definite integration is crucial when dealing with boundaries or limits. Definite integrals have the form \( \int_{a}^{b} f(x) \, dx \) and provide a numerical value representing the integral from point \( a \) to point \( b \).
In practice, definite integration requires evaluating the antiderivative at the upper and lower limits and subtracting the results. While not explicitly shown in the provided solution, the integral transformation and simplification techniques can be applied in definite integrals with additional steps to handle limits during back-substitution.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as \( \arctan \), are the inverse operations of trigonometric functions. They are used to solve angles given the value of a trigonometric ratio. In integration, these functions frequently arise when solving integrals that involve derivatives of inverse trigonometric functions or when reversing a trigonometric substitution.
In the exercise, after substituting and integrating, the solution involves expressing the angle \( \theta \) back in terms of the original variable \( y \). This leads to \( \theta = \arctan\left(\frac{y-1}{2}\right) \), which simplifies the integration result into a meaningful expression relative to the original variable. Understanding inverse trigonometric functions is essential for interpreting the final results in terms of known quantities in calculus.