Further Mathematics | Integration

🚀 Welcome to Further Pure Mathematics 3: Advanced Integration!

Hello, future mathematician! You’ve already mastered many integration techniques in previous units—from integration by parts to using partial fractions. In FP3, we unlock a powerful new set of tools focused on integrating functions related to Hyperbolic Functions and their inverses, as well as reinforcing integration leading to Inverse Trigonometric Functions.

Don't worry if this seems tricky at first. Integration is all about pattern recognition. We are simply expanding our library of standard integral "templates." By the end of this chapter, you will be able to solve integrals that look intimidating but are actually just cleverly disguised versions of these standard forms. Let’s get started!

Key Takeaway: We are learning new standard integration results involving \( \sinh x \), \( \cosh x \), \( \arcsin x \), and the newly introduced inverse hyperbolic functions (\( \operatorname{arsinh} x \), etc.).

1. Revisiting Standard Hyperbolic Integrals

We begin by reviewing the fundamental integrals of the hyperbolic functions. Remember, hyperbolic functions (\( \sinh x \), \( \cosh x \)) are often easier to integrate than their trigonometric cousins (\( \sin x \), \( \cos x \)) because the signs behave differently!

Standard Hyperbolic Results

These results are derived directly from differentiation.

• \( \int \sinh x \, dx = \cosh x + C \)
• \( \int \cosh x \, dx = \sinh x + C \)
• \( \int \operatorname{sech}^2 x \, dx = \tanh x + C \)
• \( \int \operatorname{cosech}^2 x \, dx = -\operatorname{coth} x + C \)
• \( \int \operatorname{sech} x \tanh x \, dx = -\operatorname{sech} x + C \)
• \( \int \operatorname{cosech} x \coth x \, dx = -\operatorname{cosech} x + C \)

⚠️ Common Mistake Alert: The Sign Trick!

For trigonometric functions, \( \int \cos x \, dx = \sin x \) and \( \int \sin x \, dx = -\cos x \). Notice the negative sign for the sine integral.

For hyperbolic functions, there is NO negative sign when integrating \( \sinh x \).

Simple Trick: When differentiating or integrating hyperbolic functions, the sign changes only if the original function name starts with "c" (like \( \coth x \), \( \operatorname{cosech} x \), \( \cosh x \) when differentiating).

2. Integration Leading to Inverse Trigonometric Forms

While these results might be covered in FP2 or A2, they are essential groundwork for the inverse hyperbolic forms in FP3. These formulas occur when the integrand involves quadratic expressions under a square root, or simple quadratic denominators.

The Arctan Form

This integral comes up often and relates to the derivative of \( \arctan (\frac{x}{a}) \).

\( \int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \arctan \left( \frac{x}{a} \right) + C \)

The Arcsin Form

This integral involves a square root where the constant term is dominant (\( a^2 \) minus \( x^2 \)).

\( \int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \arcsin \left( \frac{x}{a} \right) + C \quad \text{for } |x| < a \)

Analogy: Think of these standard forms as the "traffic signs" of integration. As soon as you see an expression like \( \frac{1}{9+x^2} \), your brain should immediately flash the "Arctan sign." Here, \( a=3 \).

3. Integration Leading to Inverse Hyperbolic Forms (The FP3 Focus)

This is the core new material in FP3 integration. These standard integrals result in \( \operatorname{arsinh} \), \( \operatorname{arcosh} \), or \( \operatorname{artanh} \).

Did you know? Inverse hyperbolic functions can also be expressed using logarithms (LN form). You must be comfortable with both the hyperbolic inverse notation (e.g., \( \operatorname{arsinh} \)) and the LN form, as examiners often use both.

3.1. The Arsinh Form (The "Sum of Squares" Under the Root)

This form looks similar to the arcsin form, but notice the crucial difference: it’s a sum of squares under the root, not a difference.

Standard Integral 1: The Arsinh Integral

\( \int \frac{1}{\sqrt{x^2 + a^2}} \, dx = \operatorname{arsinh} \left( \frac{x}{a} \right) + C \)

The Logarithmic Form:

\( \int \frac{1}{\sqrt{x^2 + a^2}} \, dx = \ln \left| x + \sqrt{x^2 + a^2} \right| + C \)

This is the most common form you will encounter. Since \( x^2 + a^2 \) is always positive, there are no domain restrictions on \( x \).

3.2. The Arcosh Form (The "Difference of Squares" - x Dominant)

Here, \( x^2 \) must be larger than \( a^2 \) for the square root to be defined, meaning \( |x| > a \).

Standard Integral 2: The Arcosh Integral

\( \int \frac{1}{\sqrt{x^2 - a^2}} \, dx = \operatorname{arcosh} \left( \frac{x}{a} \right) + C \quad \text{for } x > a \)

The Logarithmic Form:

\( \int \frac{1}{\sqrt{x^2 - a^2}} \, dx = \ln \left| x + \sqrt{x^2 - a^2} \right| + C \quad \text{for } x > a \)

Memory Aid: Notice how the term inside the LN forms (\( x + \sqrt{\dots} \)) is exactly the same as the denominator in the original integral, just without the square root over the whole expression.

3.3. The Artanh/Arcoth Form (Partial Fractions Revisited)

This integral involves a quadratic denominator without a square root. Depending on the order, it yields either \( \operatorname{artanh} \) or \( \operatorname{arcoth} \).

Standard Integral 3a: Artanh (Constant Dominant)

Used when the constant comes first in the denominator (\( a^2 - x^2 \)). Requires \( |x| < a \).

\( \int \frac{1}{a^2 - x^2} \, dx = \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right| + C \quad \text{for } |x| < a \)

(Note: This integral is often derived using partial fractions: \( \frac{1}{a^2 - x^2} = \frac{1}{2a} \left( \frac{1}{a+x} + \frac{1}{a-x} \right) \)).

Standard Integral 3b: Arcoth (Variable Dominant)

Used when the variable comes first in the denominator (\( x^2 - a^2 \)). Requires \( |x| > a \).

\( \int \frac{1}{x^2 - a^2} \, dx = \frac{1}{2a} \ln \left| \frac{x-a}{x+a} \right| + C \quad \text{for } |x| > a \)

4. Advanced Technique: Completing the Square to Use Standard Forms

Rarely will the exam question present the integral in the perfect form \( x^2 + a^2 \). Often, you will face a complex quadratic expression. You must use Completing the Square to transform the denominator into one of the standard FP3 forms.

Step-by-Step Process: Transforming the Integral

Let’s look at how to tackle an integral like \( I = \int \frac{1}{\sqrt{x^2 + 6x + 13}} \, dx \).

1. Complete the Square (Denominator): Focus on the quadratic part: \( x^2 + 6x + 13 \).

   \( x^2 + 6x + 13 = (x + 3)^2 - (3)^2 + 13 \)

   \( = (x + 3)^2 - 9 + 13 = (x + 3)^2 + 4 \)

2. Substitute and Recognize the Form: The integral becomes:

   \( I = \int \frac{1}{\sqrt{(x + 3)^2 + 4}} \, dx \)

   This now perfectly matches the Arsinh form: \( \int \frac{1}{\sqrt{u^2 + a^2}} \, du \).

3. Perform a Simple Substitution (if needed): Let \( u = x+3 \). Then \( du = dx \).

   Here, \( u = x+3 \) and \( a^2 = 4 \), so \( a = 2 \).

4. Apply the Standard Result:

   \( I = \int \frac{1}{\sqrt{u^2 + 2^2}} \, du = \operatorname{arsinh} \left( \frac{u}{2} \right) + C \)

5. Substitute back:

   \( I = \operatorname{arsinh} \left( \frac{x+3}{2} \right) + C \)

Important Tip: If the coefficient of \( x^2 \) is not 1 (e.g., \( 2x^2 + \dots \)), you must factor out that coefficient first before completing the square!

5. Integration Using Hyperbolic Substitution (Proof Method)

Sometimes, you may be asked to evaluate a definite integral or prove one of the standard results using a specific hyperbolic substitution. This leverages the hyperbolic identities you learned earlier.

The standard substitutions are:

• To simplify \(\sqrt{x^2 + a^2}\), use \( x = a \sinh \theta \). (Uses identity \( 1 + \sinh^2 \theta = \cosh^2 \theta \)).
• To simplify \(\sqrt{x^2 - a^2}\), use \( x = a \cosh \theta \). (Uses identity \( \cosh^2 \theta - 1 = \sinh^2 \theta \)).

Example Walkthrough: Using Substitution

Evaluate \( \int \frac{1}{\sqrt{x^2 + 9}} \, dx \) using the substitution \( x = 3 \sinh \theta \).

1. Find \( dx \) and \( \sqrt{x^2 + 9} \):

   If \( x = 3 \sinh \theta \), then \( \frac{dx}{d\theta} = 3 \cosh \theta \), so \( dx = 3 \cosh \theta \, d\theta \).

   \( \sqrt{x^2 + 9} = \sqrt{(3 \sinh \theta)^2 + 9} = \sqrt{9 (\sinh^2 \theta + 1)} \)

   Using the identity: \( = \sqrt{9 \cosh^2 \theta} = 3 \cosh \theta \)

2. Substitute into the integral:

   \( I = \int \frac{1}{3 \cosh \theta} \cdot (3 \cosh \theta \, d\theta) \)

   The \( 3 \cosh \theta \) terms cancel beautifully!

   \( I = \int 1 \, d\theta = \theta + C \)

3. Substitute back to \( x \):

   Since \( x = 3 \sinh \theta \), then \( \frac{x}{3} = \sinh \theta \), which means \( \theta = \operatorname{arsinh} (\frac{x}{3}) \).

   \( I = \operatorname{arsinh} \left( \frac{x}{3} \right) + C \) (Matching the standard result!)

6. Summary of Advanced Integration in FP3

FP3 integration is predominantly about recognizing and applying the new standard integrals related to inverse hyperbolic functions. Success relies on two main skills:

1. Memorization/Familiarity: Knowing the three main standard forms (Arsinh, Arcosh, Artanh/coth) and their corresponding logarithmic expressions.
2. Manipulation: Using completing the square or simple linear substitution (\( u = x+k \)) to convert tricky integrands into one of the standard forms.

Keep these formulas handy, review the steps for completing the square, and remember the unique sign conventions for integrating hyperbolic functions. Good luck!