Integration

💪 Pure Mathematics 3 (Paper 3) Study Notes: Integration (Topic 3.5) 💪

Hello A-Level Mathematicians! Welcome to the Integration chapter for Paper 3. If differentiation was about finding the rate of change, integration is the ultimate detective work—finding the original function (and the area under it!).

In Pure Mathematics 1 and 2, you learned the basics. Now, in P3, we encounter the truly tricky functions—products, quotients, and forms that require advanced algebraic manipulation before you can even start. This chapter is vital, as these skills are frequently used in the subsequent chapter on Differential Equations (Topic 3.8).

Ready to master the hardest integrals? Let's dive in!

1. Extending Basic Integration Forms (Reverse Differentiation)

Remember, integration is simply the reverse of differentiation. For P3, we must be comfortable integrating exponential, trigonometric, and rational functions involving the factor \((ax+b)\).

Quick Review: The \(ax+b\) Rule

When integrating a function of the form \(f(ax+b)\), we integrate \(f\) normally and then divide by the derivative of the inner function (\(a\)). Don't forget the constant of integration, \(+C\)!

• \( \int e^{ax+b} \, dx = \frac{1}{a} e^{ax+b} + C \)
• \( \int \sin(ax+b) \, dx = -\frac{1}{a} \cos(ax+b) + C \)
• \( \int \cos(ax+b) \, dx = \frac{1}{a} \sin(ax+b) + C \)
• \( \int \sec^2(ax+b) \, dx = \frac{1}{a} \tan(ax+b) + C \)
• \( \int \frac{1}{ax+b} \, dx = \frac{1}{a} \ln|ax+b| + C \)

Did you know? The \(\sec^2 x\) integral is straightforward because you know its derivative is \(\tan x\). Make sure you know all standard trigonometric derivatives for P3!

The Inverse Tangent Form (A New P3 Standard)

We are now introduced to a new standard integral that results in an inverse trigonometric function:

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

Memory Aid: Notice the symmetry. If you differentiate \(\tan^{-1}\left(\frac{x}{a}\right)\), you get \(\frac{1}{1+(\frac{x}{a})^2} \cdot \frac{1}{a}\). The extra factor of \(\frac{1}{a}\) in the integration formula is needed to compensate for this.

2. Integration using Trigonometric Identities

You cannot integrate a squared trigonometric function directly (like \(\int \sin^2 x \, dx\)) because it is effectively a product (\(\sin x \cdot \sin x\)).

The trick is to use double angle formulas to rewrite the expression so that it becomes a sum or difference of simple terms we can integrate.

Key Identity Conversions:

• To integrate \(\sin^2 x\) or \(\cos^2 x\), use the identities derived from: \( \cos 2x = 2\cos^2 x - 1 \quad \text{or} \quad \cos 2x = 1 - 2\sin^2 x \)
• Integration Formula for \(\sin^2 x\): \( \sin^2 x = \frac{1}{2}(1 - \cos 2x) \)
  Then: \( \int \sin^2 x \, dx = \int \frac{1}{2}(1 - \cos 2x) \, dx = \frac{1}{2}x - \frac{1}{4}\sin 2x + C \)
• Integration Formula for \(\cos^2 x\): \( \cos^2 x = \frac{1}{2}(1 + \cos 2x) \)
  Then: \( \int \cos^2 x \, dx = \int \frac{1}{2}(1 + \cos 2x) \, dx = \frac{1}{2}x + \frac{1}{4}\sin 2x + C \)

Common Mistake: Forgetting to divide by 2 when integrating \(\cos(2x)\) or \(\sin(2x)\). Remember the \(ax+b\) rule!

3. Integration of Rational Functions via Partial Fractions

This is one of the most powerful P3 techniques. When you see a fraction where the denominator is a polynomial that can be factorised, think Partial Fractions!

Analogy: Imagine a complicated fraction is like a complicated jigsaw puzzle piece. We break it down into several simpler pieces (partial fractions) that we know how to integrate (usually resulting in \(\ln\) terms or \(\tan^{-1}\) terms).

Step-by-Step Process:

1. Check Degree: Ensure the degree of the numerator is less than the degree of the denominator. (If not, you must perform algebraic long division first, although this is usually restricted in P3 questions).
2. Decomposition: Express the integrand as a sum of simpler fractions.
3. Solve for Constants: Use substitution or coefficient comparison to find the unknown constants (A, B, C, etc.).
4. Integrate: Integrate each partial fraction separately. These usually result in \(\ln\) terms or the \(\tan^{-1}\) form \(\frac{1}{x^2+a^2}\).

The Three Types of Denominators (from P3 Syllabus 3.1):

This technique is restricted to denominators no more complicated than:

1. Distinct Linear Factors: (e.g., \((ax+b)(cx+d)\))
   \( \frac{P(x)}{(x-1)(x+2)} \equiv \frac{A}{x-1} + \frac{B}{x+2} \)
   (Integration yields two \(\ln\) terms.)
2. Repeated Linear Factors: (e.g., \((ax+b)(cx+d)^2\))
   \( \frac{P(x)}{(x-1)(x+2)^2} \equiv \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{(x+2)^2} \)
   (The final term integrates using the power rule, treating \((x+2)^{-2}\) as \(u^n\).)
3. Linear and Irreducible Quadratic Factor: (e.g., \((ax+b)(cx^2+d)\))
   \( \frac{P(x)}{(x-1)(x^2+4)} \equiv \frac{A}{x-1} + \frac{Bx+C}{x^2+4} \)
   (The quadratic factor usually requires separating the integral into two parts: one for the \(\ln\) form, and one for the \(\tan^{-1}\) form.)

4. Recognizing the Logarithmic Form: \(\int \frac{k f'(x)}{f(x)} \, dx\)

This is an absolute necessity for P3, often used alongside partial fractions or substitution.

The Principle

If the numerator is the derivative of the denominator (or a constant multiple of it), the integral is based on the natural logarithm of the denominator.

\( \int \frac{f'(x)}{f(x)} \, dx = \ln|f(x)| + C \)

Step-by-Step Example

Find \( \int \frac{3x^2 + 2}{x^3 + 2x - 5} \, dx \).

1. Identify the denominator: \(f(x) = x^3 + 2x - 5\).
2. Find its derivative: \(f'(x) = 3x^2 + 2\).
3. Compare: The numerator IS exactly \(f'(x)\)!
4. Result: \( \ln|x^3 + 2x - 5| + C \).

Handling Constants (\(k\))

What if the derivative is missing a constant? You must adjust!

Find \( \int \frac{x}{x^2 + 1} \, dx \).

1. Denominator: \(f(x) = x^2 + 1\). Derivative: \(f'(x) = 2x\).
2. We only have \(x\) in the numerator, not \(2x\). We need to introduce the factor 2.
3. Write the integral as: \( \frac{1}{2} \int \frac{2x}{x^2 + 1} \, dx \)
4. Result: \( \frac{1}{2} \ln|x^2 + 1| + C \).

5. Integration by Parts: The Reverse Product Rule

Integration by parts is used when you need to integrate a product of two different types of functions (e.g., \(x \sin x\) or \(x^2 e^x\) or \(\ln x\)).

The Formula

If \(u\) and \(v\) are functions of \(x\):
\( \int u \frac{dv}{dx} \, dx = uv - \int v \frac{du}{dx} \, dx \)

Choosing \(u\) and \(\frac{dv}{dx}\)

The success of integration by parts depends entirely on which function you choose as \(u\) (the one to differentiate) and which you choose as \(\frac{dv}{dx}\) (the one to integrate).

The goal is to make the new integral (\(\int v \frac{du}{dx} \, dx\)) simpler than the original one.

The Selection Trick (ILATE/LIATE): Choose \(u\) according to this priority order:

• L: Logarithmic functions (e.g., \(\ln x\))
• I: Inverse trigonometric functions (e.g., \(\tan^{-1} x\))
• A: Algebraic functions (e.g., \(x, x^2\))
• T: Trigonometric functions (e.g., \(\sin x, \cos x\))
• E: Exponential functions (e.g., \(e^x\))

Example: To integrate \(x \sin 2x\). \(x\) is Algebraic (A), \(\sin 2x\) is Trigonometric (T). A comes before T in LIATE, so let \(u = x\) and \(\frac{dv}{dx} = \sin 2x\).

Integrating \(\ln x\)

You cannot integrate \(\ln x\) directly. Integration by parts is the only way!

To find \( \int \ln x \, dx \), treat the integral as a product: \( \int (\ln x \cdot 1) \, dx \).

• Let \(u = \ln x\) (Logarithmic, priority 1)
• Let \(\frac{dv}{dx} = 1\) (Algebraic)

Advanced Integration by Parts (Repeated & Cyclical)

• Repeated Application: If you integrate a function like \(x^2 e^x\), you will have to apply the formula twice because differentiating \(x^2\) only eliminates the power after two steps.
• Cyclical Integrals: Functions like \(\int e^x \sin x \, dx\) seem to repeat themselves. After two applications of the formula, the original integral reappears on the right side. You must then treat the integral itself as an unknown variable (\(I\)) and solve the resulting algebraic equation to find \(I\).

6. Integration by Substitution

In P3, you are expected to use a given substitution to simplify and evaluate an integral. This means you don't need to guess the substitution; the question provides it (e.g., "Use the substitution \(u = \sin x\)").

This technique transforms a complicated integral involving \(x\) into a simpler integral involving \(u\).

Step-by-Step Substitution Process

We want to solve \( \int f(x) \, dx \) using the substitution \( u = g(x) \).

1. Differentiate \(u\): Find \(\frac{du}{dx}\). Example: If \(u = \sin x\), then \(\frac{du}{dx} = \cos x\).
2. Substitute \(dx\): Rearrange the derivative to find an expression for \(dx\) in terms of \(du\). Example: \(dx = \frac{du}{\cos x}\).
3. Substitute \(x\) terms: Replace all remaining \(x\) terms in the original integral using the substitution definition \(u=g(x)\). All terms must be in \(u\) and \(du\).
4. (Crucial for Definite Integrals) Change Limits: If the integral is definite (has limits \(a\) and \(b\)), calculate the new limits \(u_1\) and \(u_2\) by plugging \(a\) and \(b\) into the substitution \(u = g(x)\). \( u_1 = g(a) \quad \text{and} \quad u_2 = g(b) \)
5. Integrate: Solve the simplified integral in terms of \(u\).
6. Final Step: If indefinite, substitute \(u\) back with \(g(x)\). If definite, evaluate using the new limits \(u_1\) and \(u_2\).

Analogy: Substitution is like changing currency. You can't start solving a complex problem in US Dollars and then finish it in Euros without converting all your figures (including the start and end amounts—the limits!)

7. Numerical Integration: The Trapezium Rule

When a function cannot be integrated analytically, or when data is given at discrete points, we use the Trapezium Rule to approximate the value of a definite integral.

The Formula

To estimate \( \int_a^b y \, dx \) using \(n\) strips of equal width \(h = \frac{b - a}{n}\):

\( \int_a^b y \, dx \approx \frac{1}{2}h \left[ (y_0 + y_n) + 2(y_1 + y_2 + \dots + y_{n-1}) \right] \)

• \(h\): Strip width (step size).
• \(y_0, y_n\): The "end" ordinates (first and last \(y\)-values).
• \(y_1, \dots, y_{n-1}\): The "inner" ordinates (multiplied by 2).
• Note: \(n\) strips require \(n+1\) ordinates (\(y\)-values).

Determining Over-estimate vs. Under-estimate

You are often asked in Paper 3 to state whether the trapezium rule gives an over-estimate or an under-estimate, by considering a sketch graph:

• Over-estimate: When the curve is convex downwards (the tangents lie below the curve; chord tops lie above the curve).
• Under-estimate: When the curve is concave downwards (the curve bends downwards; chord tops lie below the curve).