Indefinite integration and its applications

Mathematics M2 (Algebra and Calculus) · Senior Secondary (HKDSE) · 6 min read · Updated: 24 Aug 2026

Welcome to Indefinite Integration!

Hello! Welcome to the exciting world of indefinite integration. If differentiation was about finding the rate of change (like the speed of a car), integration is about going backward – figuring out the total distance travelled from the speed. It is a fundamental tool in calculus with tons of applications in science, engineering, and economics.

Don't worry if this seems tricky at first. We will break everything down into simple, manageable steps. Think of it as learning a new set of problem-solving tools. Ready? Let's begin!

1. What is Indefinite Integration? The Reverse of Differentiation

The simplest way to think about integration is as the opposite of differentiation. We call this process finding the antiderivative.

Remember how taking the derivative of \(x^2\) gives you \(2x\)?
\(\frac{d}{dx}(x^2) = 2x\)
Well, integrating \(2x\) takes you back to \(x^2\). We write it like this:
\(\int 2x \,dx = x^2 + C\)

Understanding the Notation

The Integral Sign (\(\int\)): This elongated 'S' shape indicates integration.
The Integrand (\(2x\)): This is the function you are integrating.
The Differential (\(dx\)): This tells us we are integrating with respect to the variable \(x\).
The Constant of Integration (\(+ C\)): An arbitrary constant representing any vertical shift.

Why the "+ C"? The Mystery of the Constant of Integration

Think about these functions:
\(y = x^2\)
\(y = x^2 + 5\)
\(y = x^2 - 100\)

What happens when you differentiate all of them?
\(\frac{d}{dx}(x^2) = 2x\)
\(\frac{d}{dx}(x^2 + 5) = 2x\)
\(\frac{d}{dx}(x^2 - 100) = 2x\)

They all have the same derivative! This is because the derivative of any constant is zero. So, when we go in reverse (integrate), we don't know what the original constant was. It could have been 5, -100, or any other number.

To account for this unknown constant, we always add \(+ C\) to our answer. This represents the entire family of antiderivatives.

Key Takeaway

Indefinite integration is the reverse of differentiation. Because the derivative of a constant is zero, we must always include a constant of integration, \(C\), in our final result.

2. Your Integration Starter Pack: Basic Rules and Formulas

Just like with differentiation, there are fundamental rules and formulas you need to master in HKDSE M2.

Basic Properties of Integrals

1. The Constant Multiple Rule: \(\int k f(x) \,dx = k \int f(x) \,dx\)
Example: \(\int 5 \cos x \,dx = 5 \int \cos x \,dx\)

2. The Sum/Difference Rule: \(\int (f(x) \pm g(x)) \,dx = \int f(x) \,dx \pm \int g(x) \,dx\)
Example: \(\int (x^2 + e^x) \,dx = \int x^2 \,dx + \int e^x \,dx\)

The Essential Standard Integrals

Power Rule: \(\int x^n \,dx = \frac{x^{n+1}}{n+1} + C\) (for any \(n \neq -1\))
Constant Rule: \(\int k \,dx = kx + C\)
The Natural Log Rule: \(\int \frac{1}{x} \,dx = \ln|x| + C\)
The Exponential Rule: \(\int e^x \,dx = e^x + C\), and more generally \(\int e^{kx} \,dx = \frac{1}{k}e^{kx} + C\)
Standard Trigonometric Integrals:
\(\int \cos x \,dx = \sin x + C\)
\(\int \sin x \,dx = -\cos x + C\)
\(\int \sec^2 x \,dx = \tan x + C\)
\(\int \csc^2 x \,dx = -\cot x + C\)
\(\int \sec x \tan x \,dx = \sec x + C\)
\(\int \csc x \cot x \,dx = -\csc x + C\)

3. Integrating Trigonometric Functions Using Identities

In HKDSE M2, integrals of trigonometric powers or products frequently appear. We simplify them using standard trigonometric identities before integrating.

Using Double-Angle Formulas

To integrate \(\sin^2 x\) or \(\cos^2 x\), rewrite them using \(\cos 2x = 1 - 2\sin^2 x = 2\cos^2 x - 1\):

\(\sin^2 x = \frac{1 - \cos 2x}{2} \implies \int \sin^2 x \,dx = \int \frac{1 - \cos 2x}{2} \,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C\)
\(\cos^2 x = \frac{1 + \cos 2x}{2} \implies \int \cos^2 x \,dx = \int \frac{1 + \cos 2x}{2} \,dx = \frac{x}{2} + \frac{\sin 2x}{4} + C\)

Using Product-to-Sum Formulas

For products like \(\sin(mx)\cos(nx)\), convert them into sums:
Example: Find \(\int \sin 3x \cos 2x \,dx\)
Using \(\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]\):
\(\int \sin 3x \cos 2x \,dx = \frac{1}{2} \int (\sin 5x + \sin x) \,dx = -\frac{\cos 5x}{10} - \frac{\cos x}{2} + C\)

4. Finding Particular Solutions (Initial Value Problems)

When an initial or boundary condition is provided, we can determine the exact value of \(C\).

Problem: The gradient of a curve is \(\frac{dy}{dx} = 3x^2 - 4x\). If the curve passes through \((2, 1)\), find its equation.

Step 1: Integrate the gradient function.
\(y = \int (3x^2 - 4x) \,dx = x^3 - 2x^2 + C\)

Step 2: Substitute \(x = 2\) and \(y = 1\) to solve for \(C\).
\(1 = (2)^3 - 2(2)^2 + C \implies 1 = 8 - 8 + C \implies C = 1\)

Step 3: State the particular solution:
Final Answer: \(y = x^3 - 2x^2 + 1\)

5. Integration by Substitution

Integration by Substitution is the reverse of the Chain Rule. It simplifies an integral by replacing an algebraic sub-expression with a single variable \(u\).

Problem: Find \(\int 2x\sqrt{x^2+1} \,dx\)

Step 1: Let \(u = x^2+1\).
Step 2: Find \(du = 2x \,dx\).
Step 3: Substitute into the integral:
\(\int \sqrt{u} \,du = \int u^{1/2} \,du = \frac{u^{3/2}}{3/2} + C = \frac{2}{3}u^{3/2} + C\)
Step 4: Back-substitute \(u = x^2+1\):
Final Answer: \(\frac{2}{3}(x^2+1)^{3/2} + C\)

6. Trigonometric Substitution

Trigonometric substitution is used for expressions containing radicals of quadratic forms or sums of squares.

Two Standard Forms for HKDSE M2:

1. For \(\sqrt{a^2 - x^2}\) or \(\frac{1}{\sqrt{a^2 - x^2}}\):
Use \(x = a \sin\theta\), where \(dx = a \cos\theta \,d\theta\).
Identity used: \(a^2 - a^2\sin^2\theta = a^2\cos^2\theta\).

2. For \(x^2 + a^2\) or \(\frac{1}{x^2 + a^2}\):
Use \(x = a \tan\theta\), where \(dx = a \sec^2\theta \,d\theta\).
Identity used: \(a^2\tan^2\theta + a^2 = a^2\sec^2\theta\).

Standard Result:

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

7. Integration by Parts

Integration by parts reverses the product rule for differentiation. It applies to products of two distinct function types.

Formula:

\(\int u \,dv = uv - \int v \,du\)

Choosing \(u\) using the LIATE Rule:

Logarithmic (e.g., \(\ln x\))
Inverse trigonometric (e.g., \(\tan^{-1} x\))
Algebraic (e.g., \(x, x^2\))
Trigonometric (e.g., \(\sin x, \cos x\))
Exponential (e.g., \(e^x\))

Example:

Problem: Find \(\int x \cos x \,dx\)

Let \(u = x\) and \(dv = \cos x \,dx\).
Then \(du = dx\) and \(v = \sin x\).
Applying the formula:
\(\int x \cos x \,dx = x \sin x - \int \sin x \,dx = x \sin x - (-\cos x) + C = x \sin x + \cos x + C\)

Note for HKDSE: Under the M2 syllabus, questions require applying integration by parts at most twice in a single problem.

References

Mathematics Education - Curriculum Documents — Education Bureau (EDB), Hong Kong SAR Government
Curriculum and Assessment Guide (Secondary 4-6) Mathematics — Curriculum Development Council & Hong Kong Examinations and Assessment Authority (HKEAA)

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is indefinite integration and why must we include the constant of integration \(C\)?

Indefinite integration is the reverse process of differentiation (finding the antiderivative), and \(+ C\) is added because the derivative of any constant is zero.

What are the standard rules for integrating powers, constants, \(\frac{1}{x}\), and exponentials?

\(\int x^n \,dx = \frac{x^{n+1}}{n+1} + C\) \((n \neq -1)\), \(\int k \,dx = kx + C\), \(\int \frac{1}{x} \,dx = \ln|x| + C\), and \(\int e^{kx} \,dx = \frac{1}{k}e^{kx} + C\).

How do you integrate even powers of basic trigonometric functions such as \(\sin^2 x\) and \(\cos^2 x\)?

Use the double-angle identities \(\sin^2 x = \frac{1 - \cos 2x}{2}\) and \(\cos^2 x = \frac{1 + \cos 2x}{2}\) to reduce them to linear cosine terms before integrating.

How do you integrate products of sines and cosines with different frequencies such as \(\sin(mx)\cos(nx)\)?

Convert the product into a sum or difference of sine and cosine terms using product-to-sum trigonometric identities.

How do you find a particular solution (initial value problem) from a given derivative?

Integrate the derivative to obtain the general solution with \(+ C\), then substitute the given initial point \((x, y)\) to solve for the specific value of \(C\).

What is integration by substitution and when should it be used?

It is the reverse of the Chain Rule, used by setting \(u = g(x)\) so that \(f(g(x))g'(x)\,dx\) simplifies to \(\int f(u)\,du\).

What substitutions are used for integrals containing \(\sqrt{a^2 - x^2}\) and \(x^2 + a^2\)?

Use \(x = a\sin\theta\) for \(\sqrt{a^2 - x^2}\), and \(x = a\tan\theta\) for \(x^2 + a^2\) (giving \(\int \frac{1}{x^2+a^2}\,dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C\)).

What is the formula for integration by parts and how do you choose \(u\)?

The formula is \(\int u \,dv = uv - \int v \,du\), and \(u\) is chosen according to the LIATE rule (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential).

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