Mathematics - Further (9231) | Integration

Welcome to Integration in Further Pure Mathematics 2!

Integration in FP2 is where we take the techniques you mastered in A Level Mathematics (9709) and apply them to tougher, more exotic functions and complex geometric problems. Don't worry if previous integration felt challenging—we will break down these advanced methods, like Reduction Formulae and geometric applications (Arc Lengths and Surface Areas), into manageable steps.

This chapter is essential for connecting pure mathematics with geometry and physics, giving you powerful tools to solve real-world problems involving curves and volumes. Let’s dive in!

1. Integration of Hyperbolic Functions

Since you have already studied Hyperbolic Functions (Topic 2.1), integration simply involves reversing the differentiation rules. Knowing your derivatives is key!

1.1 Standard Results

Recall that the derivatives of hyperbolic functions are very similar to trigonometric functions, but watch the signs!

• \(\int \sinh x \, dx = \cosh x + C\)
• \(\int \cosh x \, dx = \sinh x + C\)
• \(\int \sech^2 x \, dx = \tanh x + C\)
• \(\int \sech x \tanh x \, dx = -\sech x + C\) (Note the minus sign reversal compared to differentiation)

1.2 Integrating Powers and Products

Just like ordinary trigonometric integration, when dealing with powers of hyperbolic functions, you often need to use identities or substitution.

The fundamental identity is: \(\cosh^2 x - \sinh^2 x = 1\).

Example: Integrating \(\sinh^2 x\)
To integrate \(\sinh^2 x\), we can't do it directly. We use the double angle identity (similar to \(\cos 2x\)): $$ \cosh 2x = \cosh^2 x + \sinh^2 x $$ Since \(\cosh^2 x = 1 + \sinh^2 x\), substituting this gives: $$ \cosh 2x = (1 + \sinh^2 x) + \sinh^2 x = 1 + 2\sinh^2 x $$ Rearranging for \(\sinh^2 x\): $$ \sinh^2 x = \frac{1}{2}(\cosh 2x - 1) $$ Now, integration becomes straightforward: $$ \int \sinh^2 x \, dx = \int \frac{1}{2}(\cosh 2x - 1) \, dx = \frac{1}{2}\left(\frac{\sinh 2x}{2} - x\right) + C $$

2. Inverse Trigonometric and Hyperbolic Integrals

In FP2, you must be able to recognise and integrate three special forms that result in inverse trigonometric or inverse hyperbolic functions.

Did you know? These integrals are crucial because they naturally arise when finding the arc length of complex curves!

2.1 The Three Key Standard Forms

You need to be able to integrate functions of the form:

1. \(\frac{1}{\sqrt{a^2 - x^2}}\) (Inverse Sine Form):
   This form contains the square root of (constant squared - variable squared). It leads to inverse trigonometric results. $$ \int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \sin^{-1}\left(\frac{x}{a}\right) + C $$
2. \(\frac{1}{\sqrt{x^2 + a^2}}\) (Inverse Hyperbolic Sine Form):
   This form contains the square root of (variable squared + constant squared). $$ \int \frac{1}{\sqrt{x^2 + a^2}} \, dx = \sinh^{-1}\left(\frac{x}{a}\right) + C $$
3. \(\frac{1}{\sqrt{x^2 - a^2}}\) (Inverse Hyperbolic Cosine Form):
   This form contains the square root of (variable squared - constant squared). (Restricted to \(|x| > a\)). $$ \int \frac{1}{\sqrt{x^2 - a^2}} \, dx = \cosh^{-1}\left(\frac{x}{a}\right) + C $$

Analogy Aid: Think of the sign in the denominator:

• If \(a^2\) is positive and \(x^2\) is negative (\(a^2 - x^2\)): It’s Sine (Standard Trig).
• If \(x^2\) is positive (\(x^2 + a^2\) or \(x^2 - a^2\)): It’s Hyperbolic.

2.2 The Essential Technique: Completing the Square

Often, the integral you receive will not immediately look like one of the standard forms. You must first use completing the square on the expression inside the square root.

Process Example: Integrate \(\int \frac{1}{\sqrt{x^2 + 6x + 13}} \, dx\).

1. Complete the Square: Focus on the quadratic \(x^2 + 6x + 13\). $$ x^2 + 6x + 13 = (x^2 + 6x + 9) + 13 - 9 = (x + 3)^2 + 4 $$
2. Substitute and Adjust: Rewrite the integral using the completed square form. $$ \int \frac{1}{\sqrt{(x + 3)^2 + 4}} \, dx $$
3. Recognise the Form: This matches Form 2: \(\int \frac{1}{\sqrt{u^2 + a^2}} \, du\), where \(u = x + 3\) and \(a = 2\). Since the substitution \(u = x+3\) means \(du = dx\), no extra factor is needed.
4. Final Integration: Use the standard result. $$ \sinh^{-1}\left(\frac{u}{a}\right) + C = \sinh^{-1}\left(\frac{x+3}{2}\right) + C $$

3. Reduction Formulae

Integration is often difficult when the integrand contains a high power, such as \(\int x^4 e^x \, dx\) or \(\int \cos^5 x \, dx\). A Reduction Formula allows you to express an integral involving a power \(n\) in terms of a similar integral involving a lower power (like \(n-1\) or \(n-2\)). This "reduces" the difficulty until you reach a power you can easily evaluate (like $n=0$ or $n=1$).

3.1 The Derivation Method: Integration by Parts (IBP)

The vast majority of reduction formulae are derived using the technique of Integration by Parts (IBP), perhaps applied multiple times.

Recall IBP: \(\int u \, \frac{dv}{dx} \, dx = uv - \int v \, \frac{du}{dx} \, dx\).

Step-by-Step Derivation Process

Let \(I_n\) be the integral you are trying to simplify (e.g., \(I_n = \int \sin^n x \, dx\)).

1. Split the Integrand: Separate the integrand into two parts, \(u\) and \(\frac{dv}{dx}\), such that when you apply IBP, the resulting integral contains a lower power.
   For \(I_n = \int \sin^n x \, dx\), we often split it as: $$ I_n = \int \sin^{n-1} x \cdot (\sin x) \, dx $$
2. Apply IBP: Choose \(u\) and \(\frac{dv}{dx}\). This requires strategic thinking!
   If you choose \(u = \sin^{n-1} x\) and \(\frac{dv}{dx} = \sin x\):
   • \(v = -\cos x\)
   • \(\frac{du}{dx} = (n-1)\sin^{n-2} x \cos x\)
   Applying IBP: $$ I_n = \sin^{n-1} x (-\cos x) - \int (-\cos x) (n-1)\sin^{n-2} x \cos x \, dx $$
3. Simplify and Substitute Identities: You will usually use an identity (like \(\sin^2 x + \cos^2 x = 1\)) to relate the new integral back to the original form \(I_{n-2}\) or \(I_n\). $$ I_n = -\cos x \sin^{n-1} x + (n-1) \int \sin^{n-2} x \cos^2 x \, dx $$ Replace \(\cos^2 x\) with \(1 - \sin^2 x\): $$ I_n = -\cos x \sin^{n-1} x + (n-1) \int \sin^{n-2} x (1 - \sin^2 x) \, dx $$ $$ I_n = -\cos x \sin^{n-1} x + (n-1) \left[ \int \sin^{n-2} x \, dx - \int \sin^n x \, dx \right] $$ $$ I_n = -\cos x \sin^{n-1} x + (n-1) I_{n-2} - (n-1) I_n $$
4. Isolate \(I_n\): Solve the resulting equation to express \(I_n\) in terms of \(I_{n-2}\) (the reduction formula). $$ I_n + (n-1) I_n = -\cos x \sin^{n-1} x + (n-1) I_{n-2} $$ $$ n I_n = -\cos x \sin^{n-1} x + (n-1) I_{n-2} $$ This is the required reduction formula.

3.2 Using Reduction Formulae

Once derived, the formula is used iteratively to find the value of a definite integral.

Example: Evaluate \(\int_0^{\pi/2} \sin^4 x \, dx\).
If \(J_n = \int_0^{\pi/2} \sin^n x \, dx\), the formula simplifies greatly since the boundary term \([-\cos x \sin^{n-1} x]_0^{\pi/2}\) evaluates to zero.

The reduction formula for definite integrals becomes: $$ J_n = \frac{n-1}{n} J_{n-2} $$

We start with \(n=4\): $$ J_4 = \frac{3}{4} J_2 $$ Next, we find \(J_2\): $$ J_2 = \frac{1}{2} J_0 $$ We must calculate \(J_0 = \int_0^{\pi/2} \sin^0 x \, dx = \int_0^{\pi/2} 1 \, dx = [x]_0^{\pi/2} = \frac{\pi}{2}\).

Working backwards: $$ J_2 = \frac{1}{2} \left(\frac{\pi}{2}\right) = \frac{\pi}{4} $$ $$ J_4 = \frac{3}{4} J_2 = \frac{3}{4} \left(\frac{\pi}{4}\right) = \frac{3\pi}{16} $$

4. Geometric Applications of Integration

Integration can be used to calculate lengths along a curve (Arc Length) and the surface area generated when a curve is revolved around an axis (Surface Area of Revolution).

4.1 Arc Length

Arc length measures the distance travelled along a curve between two points. We use different formulae depending on how the curve is defined.

A. Cartesian Coordinates (\(y = f(x)\))

The length \(L\) of the curve from \(x=a\) to \(x=b\) is: $$ L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx $$

B. Parametric Coordinates (\(x = x(t), y = y(t)\))

If the curve is defined parametrically by parameter \(t\) from \(t_1\) to \(t_2\): $$ L = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt $$

C. Polar Coordinates (\(r = f(\theta)\))

If the curve is defined in polar coordinates from \(\theta=\alpha\) to \(\theta=\beta\): $$ L = \int_{\alpha}^{\beta} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \, d\theta $$

Crucial Pre-step: Before integrating, you must find the derivatives (\(\frac{dy}{dx}\), \(\frac{dx}{dt}\), or \(\frac{dr}{d\theta}\)), square them, and simplify the expression under the square root as much as possible. This simplification often leads to a perfect square, allowing you to easily integrate the resulting expression.

4.2 Surface Area of Revolution

This calculates the area of the 3D surface created when a 2D curve is rotated around an axis. We use the formula for Arc Length but multiply by \(2\pi \times (\text{radius})\).

The formula is: \(\text{Area} = \int 2\pi (\text{radius}) \, dL\)

A. Rotation about the x-axis

The radius of rotation is the distance from the x-axis, which is \(y\). $$ S_x = \int_a^b 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx $$ (If defined parametrically, replace \(y\) with \(y(t)\) and \(\sqrt{1 + (\frac{dy}{dx})^2} \, dx\) with \(\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \, dt\).)

B. Rotation about the y-axis

The radius of rotation is the distance from the y-axis, which is \(x\). $$ S_y = \int_a^b 2\pi x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx $$ (Again, use the parametric form if appropriate.)

Note: You are required to calculate surface areas of revolution only for curves given in Cartesian or Parametric form. Polar surface area calculations are excluded from this syllabus.

5. Approximating Integrals, Bounds, and Deriving Inequalities

This topic links the geometric idea of area approximation (using rectangles, known as Riemann sums) to algebraic inequalities involving sums.

5.1 Area Approximation using Rectangles

The definite integral \(\int_a^b f(x) \, dx\) represents the area under the curve. We can approximate this area by summing the areas of many thin rectangles.

• If the function \(f(x)\) is increasing over the interval, using the left endpoint of the interval for the height gives an underestimate, and using the right endpoint gives an overestimate.
• If the function \(f(x)\) is decreasing, the roles are reversed.

By placing rectangles strategically (either inside or outside the curve), we can establish strict bounds for the exact integral value.

5.2 Deriving Inequalities

You are expected to use this concept to derive inequalities or limits concerning sums (series). This often involves comparing a discrete sum, \(\sum_{r=a}^b f(r)\), with the continuous integral, \(\int_a^b f(x) \, dx\).

Example: Harmonic Series and Logarithms
Consider the function \(f(x) = \frac{1}{x}\), which is decreasing for \(x > 0\). We want to relate the sum \(S_n = \sum_{r=1}^n \frac{1}{r}\) to the integral \(\int \frac{1}{x} \, dx = \ln x\).

If we consider the sum of rectangles from \(r=1\) to \(r=n\) (with unit width, \(\delta x = 1\)):

• Upper Bound: If we use the left endpoint of each interval \([r, r+1]\), the rectangle height is \(f(r) = \frac{1}{r}\). $$ \sum_{r=1}^n \frac{1}{r} > \int_1^{n+1} \frac{1}{x} \, dx $$ $$ \sum_{r=1}^n \frac{1}{r} > [\ln x]_1^{n+1} = \ln(n+1) - \ln(1) = \ln(n+1) $$
• Lower Bound: If we use the right endpoint of each interval, the height is \(f(r+1)\). This sum is slightly smaller than the total sum \(S_n\). $$ \sum_{r=1}^{n-1} \frac{1}{r+1} < \int_1^n \frac{1}{x} \, dx $$ If we look at the sum \(S_n\) and split off the first term (\(1\)): $$ S_n = 1 + \sum_{r=2}^n \frac{1}{r} $$ Using the integral \(\int_1^n \frac{1}{x} \, dx = \ln n\): $$ \sum_{r=2}^n \frac{1}{r} < \int_1^n \frac{1}{x} \, dx = \ln n $$ Therefore, \(S_n = 1 + \sum_{r=2}^n \frac{1}{r} < 1 + \ln n\).

Combining these bounds gives the standard inequality related to the harmonic series: $$ \ln(n+1) < \sum_{r=1}^n \frac{1}{r} < 1 + \ln n $$