Problem 71 - Thomas Calculus in SI Units

Question

Evaluate the integrals. $$\int \frac{d y}{y^{2}-2 y+5}$$

Step-by-Step Solution

Answer

The integral is \(\frac{1}{2} \arctan\left(\frac{y - 1}{2}\right) + C\).

Step 1: Recognize the need for completing the square

The function in the denominator, \(y^2 - 2y + 5\), is not immediately factorizable in a form that we can integrate easily. However, we notice that we can rewrite it by completing the square.

Step 2: Complete the square

Rewrite \(y^2 - 2y + 5\) as \((y - 1)^2 + 4\). We complete the square by identifying the expression that can be written in the form \((y - a)^2 + b\). For this expression: \[y^2 - 2y +1 - 1 + 5 = (y - 1)^2 + 4. \]

Step 3: Substitute to simplify the integrand

Let \(u = y - 1\), which implies \(dy = du\). Then, our integral becomes: \[\int \frac{du}{u^2 + 4}. \]

Step 4: Use the inverse trigonometric integral formula

Identify the integral as one that matches the form \(\int \frac{du}{u^2 + a^2} = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C\). Here, \(a^2 = 4\), or \(a = 2\). So, \[\int \frac{du}{u^2 + 4} = \frac{1}{2} \arctan\left(\frac{u}{2}\right) + C. \]

Step 5: Substitute back to original variable

Since \(u = y - 1\), substitute back to get the expression in terms of \(y\):\[\frac{1}{2} \arctan\left(\frac{y - 1}{2}\right) + C.\]

Key Concepts

Completing the Square

Completing the square is a crucial technique in algebra that allows us to rewrite quadratic expressions so they become easier to manipulate and integrate. Given a quadratic expression like \( y^2 - 2y + 5 \), it's often beneficial to transform it into a more convenient form. This form is

• \( (y - a)^2 + b \)

where \( a \) and \( b \) are constants.
To complete the square for an expression like \( y^2 - 2y + 5 \), we follow these steps:

• Identify the middle term, \(-2y\), and use it to find \( a \) so that \( 2a = -2 \), leading to \( a = 1 \).
• Rearrange to obtain \( (y - 1)^2 \), which expands to \( y^2 - 2y + 1 \).
• Add and subtract \( 1 \) within the original expression, ensuring it remains equivalent: \( y^2 - 2y + 5 = (y - 1)^2 + (5 - 1) = (y - 1)^2 + 4 \).

Completing the square transforms complex quadratics into a manageable form, simplifying further calculations.

Inverse Trigonometric Functions

Inverse trigonometric functions are vital for integrating certain expressions. They allow us to express angles when given ratios, which is particularly useful when dealing with integrals involving squares. In calculus, a common function used is the arctangent function, \( \arctan(x) \), especially in integrals like:

• \( \int \frac{du}{u^2 + a^2} = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C \)

In the case of our integral where we have \( \int \frac{du}{u^2 + 4} \), the inverse trigonometric formula applies perfectly.

• Here, \( a^2 = 4 \), which means \( a = 2 \).
• Thus, the integral becomes \( \frac{1}{2} \arctan\left(\frac{u}{2}\right) + C \).

The arctan function is derived from the need to express angles whose tangents we know, which often arise in contexts involving circles or rotations. This technique turns potentially complex polynomial fractions into manageable forms by bridging them with geometric interpretations.

Substitution Method

The substitution method is a useful tool that simplifies integration by changing variables. By substituting a new variable, we can transform a difficult integral into a simpler one. For instance, consider substituting \( u = y - 1 \). This step allows us to align our integral with a familiar format. By substituting, we can:

• Simplify expressions from \( u^2 + 4 \) instead of \( y^2 - 2y + 5 \).
• Ensure that \( dy = du \), maintaining the integrity of the integral's differential.

The original variable \( y \) is replaced with \( u + 1 \) due to the substitution \( u = y - 1 \). This change turns the integral into a simpler format \( \int \frac{du}{u^2 + 4} \), which is much easier to handle using inverse trigonometric functions.
By choosing an effective substitution, the integral retains all characteristics of the original problem while simplifying the mathematical operations needed to solve it.