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's a fundamental tool in calculus with tons of applications in science, engineering, and even economics.

Don't worry if this seems tricky at first. We'll 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 long 'S' shape means "integrate".
  • 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 ($$+ C$$): This is super important! We'll talk about it next.

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 functions that have the same derivative.

Key Takeaway

Indefinite integration is the reverse of differentiation. Its goal is to find the original function (the antiderivative). Because the derivative of a constant is zero, we must always add a constant of integration, C, to our result.


2. Your Integration Starter Pack: Basic Rules and Formulas

Just like with differentiation, there are some fundamental rules and formulas you need to master. Once you know these, you can solve a huge number of problems!

Basic Properties of Integrals

1. The Constant Multiple Rule: You can pull a constant multiplier out of the integral.
Example: $$ \int 5 \cos(x) \,dx = 5 \int \cos(x) \,dx $$

2. The Sum/Difference Rule: You can integrate a function term by term.
Example: $$ \int (x^2 + e^x) \,dx = \int x^2 \,dx + \int e^x \,dx $$

The Essential Formulas (from the syllabus)

These are your must-know formulas. You should aim to memorise them!

Quick Review: Integration Formulas
  • Power Rule: $$ \int x^n \,dx = \frac{x^{n+1}}{n+1} + C $$ (for any n ≠ -1)
    (Tip: Add one to the power, then divide by the new power.)
  • Constant Rule: $$ \int k \,dx = kx + C $$
    (Example: $$ \int 7 \,dx = 7x + C $$)
  • The Natural Log Rule: $$ \int \frac{1}{x} \,dx = \ln|x| + C $$
    (We use absolute value |x| because you can't take the log of a negative number!)
  • The Exponential Rule: $$ \int e^x \,dx = e^x + C $$
    (The easy one! It's its own integral.)
  • Trigonometric Rules:
    $$ \int \cos(x) \,dx = \sin(x) + C $$
    $$ \int \sin(x) \,dx = -\cos(x) + C $$
    (Memory Aid: Integrating to a "co-" function, like cos(x), often involves a negative sign.)
    $$ \int \sec^2(x) \,dx = \tan(x) + C $$
    (This is the direct reverse of differentiating tan(x)!)
Common Mistake to Avoid!

For the power rule, a common mistake is to multiply by the new power instead of dividing. Remember:
Differentiation: Power comes down and gets smaller. $$ \frac{d}{dx}(x^3) = 3x^2 $$
Integration: Power gets bigger and you divide. $$ \int x^2 \,dx = \frac{x^3}{3} + C $$

Key Takeaway

Mastering the basic integration formulas is crucial. Practice applying the Power Rule, Log Rule, Exponential Rule, and the three basic Trig Rules. Remember to always add "+ C"!


3. Finding the Original Blueprint: Applications of Indefinite Integration

So, what's the point of finding the "+ C"? In many real-world problems, we have extra information that allows us to find the exact value of C. This is like being given the slope of a road everywhere, plus one specific point's altitude, allowing you to map the entire road's height.

This is called finding a particular solution. A classic application is finding the equation of a curve given its gradient function ($$dy/dx$$) and a point it passes through.

Step-by-Step Guide to Finding a Particular Solution

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

Step 1: Integrate the gradient function to find the general equation.
The equation of the curve is $$ y = \int (3x^2 - 4x) \,dx $$.
$$ y = \frac{3x^3}{3} - \frac{4x^2}{2} + C $$ $$ y = x^3 - 2x^2 + C $$ This is the "general solution" - it represents the whole family of curves with that gradient.

Step 2: Use the given point to find the value of C.
We know the curve passes through (2, 1). This means when $$x=2$$, $$y=1$$. Let's substitute these values into our general solution.
$$ 1 = (2)^3 - 2(2)^2 + C $$ $$ 1 = 8 - 2(4) + C $$ $$ 1 = 8 - 8 + C $$ $$ C = 1 $$

Step 3: Write down the final particular solution.
Now that we know C=1, we can write the specific equation for this curve.
Final Answer: $$ y = x^3 - 2x^2 + 1 $$

Key Takeaway

Indefinite integration gives a general solution ($$y = F(x) + C$$). If you are given a point ($$x_0, y_0$$) on the curve, you can substitute it into the general solution to find the specific value of C and get the particular solution.


4. The Art of Disguise: Integration by Substitution

What happens when you have an integral that doesn't fit any of our basic formulas? Like $$ \int 2x\sqrt{x^2+1} \,dx $$?

This is where Integration by Substitution comes in. It's the reverse of the Chain Rule for differentiation. The main idea is to simplify a complicated integral by substituting part of the expression with a new variable, usually 'u'.

When to Use It?

Look for an integral where you can see a function (the "inside part") and its derivative (or a constant multiple of its derivative) also present.
In our example, $$ \int 2x\sqrt{x^2+1} \,dx $$, the "inside part" is $$x^2+1$$. Its derivative is $$2x$$, which is also in the integral! This is a perfect candidate for substitution.

Step-by-Step Guide to Substitution

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

Step 1: Choose your 'u'. Let 'u' be the "inside function".
Let $$ u = x^2+1 $$

Step 2: Find 'du'. Differentiate 'u' with respect to 'x' ($$du/dx$$) and rearrange.
$$ \frac{du}{dx} = 2x $$ Rearranging gives: $$ du = 2x \,dx $$

Step 3: Substitute everything. Replace the parts of the original integral with 'u' and 'du'. Your new integral should only have 'u' as the variable.
Our integral was $$ \int \sqrt{x^2+1} \cdot (2x \,dx) $$ Substituting gives: $$ \int \sqrt{u} \,du $$ Look how much simpler that is!

Step 4: Integrate with respect to 'u'. Use the power rule ($$\sqrt{u} = u^{1/2}$$).
$$ \int u^{1/2} \,du = \frac{u^{1/2 + 1}}{1/2 + 1} + C = \frac{u^{3/2}}{3/2} + C = \frac{2}{3}u^{3/2} + C $$

Step 5: Back-substitute. The original problem was in 'x', so our answer must be in 'x'. Replace 'u' with its original expression.
Final Answer: $$ \frac{2}{3}(x^2+1)^{3/2} + C $$

Common Mistake to Avoid!

Students often forget Step 5! You must always substitute back to the original variable at the end. An answer left in terms of 'u' is incomplete.

Key Takeaway

Integration by Substitution (u-sub) simplifies integrals by swapping out a complex part for a single variable 'u'. Look for a function and its derivative. Follow the 5 steps: Choose u, Find du, Substitute, Integrate, Back-substitute.


5. Trigonometry to the Rescue: Trigonometric Substitution

This is a special, powerful type of substitution used for integrals containing specific forms involving square roots or sums of squares. The goal is to use trigonometric identities like $$ \sin^2\theta + \cos^2\theta = 1 $$ to get rid of the tricky part.

You only need to recognise two main forms for the HKDSE M2 syllabus:

The Substitution Recipes

1. For integrals involving $$ \sqrt{a^2 - x^2} $$ or $$ \frac{1}{\sqrt{a^2 - x^2}} $$ :
Use the substitution: $$ x = a \sin\theta $$
This works because $$ a^2 - x^2 = a^2 - a^2\sin^2\theta = a^2(1-\sin^2\theta) = a^2\cos^2\theta $$. The square root disappears!

2. For integrals involving $$ x^2 + a^2 $$ or $$ \frac{1}{x^2 + a^2} $$ :
Use the substitution: $$ x = a \tan\theta $$
This works because $$ x^2 + a^2 = a^2\tan^2\theta + a^2 = a^2(\tan^2\theta + 1) = a^2\sec^2\theta $$.

Did you know? The answers to these types of integrals often involve inverse trigonometric functions like $$ \sin^{-1}(x) $$ (arcsin) or $$ \tan^{-1}(x) $$ (arctan). Let's see how.

Example: The $$1/(x^2 + a^2)$$ case

Problem: Find $$ \int \frac{1}{x^2+a^2} \,dx $$

1. Substitute: This matches our second recipe. Let $$ x = a \tan\theta $$. Then we must also find $$dx$$. Differentiating gives $$ \frac{dx}{d\theta} = a \sec^2\theta $$, so $$ dx = a \sec^2\theta \,d\theta $$.

2. Simplify: Plug these into the integral.
$$ \int \frac{1}{(a \tan\theta)^2 + a^2} \cdot (a \sec^2\theta \,d\theta) = \int \frac{1}{a^2\tan^2\theta + a^2} \cdot (a \sec^2\theta \,d\theta) $$ $$ = \int \frac{1}{a^2(\tan^2\theta + 1)} \cdot (a \sec^2\theta \,d\theta) = \int \frac{1}{a^2\sec^2\theta} \cdot (a \sec^2\theta \,d\theta) $$ The $$a\sec^2\theta$$ terms cancel out beautifully! $$ = \int \frac{1}{a} \,d\theta $$

3. Integrate: This is now a very simple integral.
$$ \frac{1}{a} \theta + C $$

4. Back-substitute: We need to express $$\theta$$ in terms of $$x$$. From our original substitution, $$ x = a\tan\theta $$, which means $$ \tan\theta = \frac{x}{a} $$. Therefore, $$ \theta = \tan^{-1}\left(\frac{x}{a}\right) $$.
Final Answer: $$ \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C $$

Key Takeaway

Trigonometric substitution is a special technique for specific forms. Memorise the two "recipes": use $$x=a\sin\theta$$ for $$a^2-x^2$$ forms, and $$x=a\tan\theta$$ for $$x^2+a^2$$ forms.


6. Divide and Conquer: Integration by Parts

Our final major technique is Integration by Parts. This is the integration version of the Product Rule for differentiation. It's used when you need to integrate a product of two different types of functions, like $$ \int x e^x \,dx $$ (an algebraic function times an exponential one).

The Magic Formula

The formula looks a bit intimidating, but you'll get used to it:

$$ \int u \,dv = uv - \int v \,du $$

The trick is to split your integral into two parts: a 'u' part (which you will differentiate) and a 'dv' part (which you will integrate).

How to Choose 'u' and 'dv'

Choosing the right 'u' is the most important step! A good rule of thumb is the acronym LIATE:

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

Whichever function type from your product appears first on this list, choose that as your 'u'.

Step-by-Step Guide to Integration by Parts

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

Step 1: Choose 'u' and 'dv'.
Our integral is a product of Algebraic ($$x$$) and Trigonometric ($$\cos(x)$$). In LIATE, 'A' comes before 'T'.
So, let $$u = x$$.
Everything else is 'dv'. So, $$dv = \cos(x) \,dx$$.

Step 2: Find 'du' and 'v'.
Differentiate 'u': $$ \frac{du}{dx} = 1 \implies du = dx $$.
Integrate 'dv': $$ v = \int \cos(x) \,dx = \sin(x) $$. (No need for '+C' here).

Step 3: Plug everything into the formula: $$ \int u \,dv = uv - \int v \,du $$
$$ \int x \cos(x) \,dx = (x)(\sin(x)) - \int (\sin(x))(dx) $$

Step 4: Solve the remaining integral.
The new integral, $$ \int \sin(x) \,dx $$, is a simple one we already know!
$$ = x\sin(x) - (-\cos(x)) + C $$
Final Answer: $$ x\sin(x) + \cos(x) + C $$

Key Takeaway

Use Integration by Parts for products of functions. Choose 'u' using the LIATE rule. Apply the formula $$ \int u \,dv = uv - \int v \,du $$. The goal is to make the new integral, $$ \int v \,du $$, simpler than the original. The syllabus states you'll only ever have to apply this process at most twice for one problem.