Ultimate Guide to Integration by Substitution in Extension 2

Introduction

In Extension 1, you were introduced to Integration by Substitution (also known as $u$ -substitution or change of variables), typically with the exact substitution provided for you in the question. In HSC Mathematics Extension 2, the training wheels come off. You will be expected to identify the correct substitution yourself, handle complex algebraic manipulations to change the integrand, and flawlessly update the bounds for definite integrals.

Executive Summary

This guide covers the mechanics of Integration by Substitution:

The Concept: Reversing the Chain Rule by substituting a complex expression with a single variable ( $u$ ).
The Differential ( $du$ ): Properly calculating and substituting $du = u'(x)dx$ .
Definite Integrals: Changing the upper and lower limits of integration so you never have to convert back to $x$ .
Finding the Substitution: Strategies for identifying the best substitution when one is not provided.

What is this about?

Integration by substitution is the integral calculus equivalent of the Chain Rule from differential calculus. When a function has an "inner" function tucked inside a power, a root, or a trigonometric function, it is difficult to integrate directly.

By defining that "inner" function as a new variable (usually $u$ ), you can transform the entire integral from the $x$ -world into the $u$ -world. If chosen correctly, the new integral will be a standard, recognizable form (like a simple polynomial or basic trig function) that is easy to evaluate. The challenge in Extension 2 is recognizing which part of the function should become $u$ .

Main Content

1. The Core Mechanics

Every substitution problem follows the same four steps:

Choose $u$ : Let $u = f(x)$ .
Find the differential $du$ : Differentiate $u$ with respect to $x$ to get $\frac{du}{dx} = f'(x)$ . Rearrange this to find $dx$ in terms of $du$ : $dx = \frac{du}{f'(x)}$ .
Substitute: Replace all instances of $x$ and $dx$ in the original integral with your new $u$ and $du$ expressions. The $x$ terms should completely cancel out.
Integrate: Evaluate the new integral with respect to $u$ .
Back-substitute (Indefinite only): If the integral is indefinite (no bounds), replace $u$ with the original $f(x)$ expression in your final answer.

2. Definite Integrals and Changing Bounds

For definite integrals, there is a crucial additional step. The bounds on the integral symbol ( $\int_a^b$ ) are $x$ -values. When you change the integral to the $u$ -world, you MUST also change the bounds to $u$ -values.

How to change bounds: If the lower bound is $x = a$ , the new lower bound is $u = f(a)$ . If the upper bound is $x = b$ , the new upper bound is $u = f(b)$ .

The golden rule of definite substitution: Once you have changed the integral and the bounds into $u$ , you never need to go back to $x$ . You just evaluate the integral using the new $u$ -bounds. This is vastly faster and less prone to algebra errors than back-substituting.

3. Choosing the Substitution

When a substitution is not given, how do you find it? Look for a function whose derivative is also present (or mostly present) in the integrand.

Common patterns:

Powers and Roots: If you have $x(x^2 + 1)^5$ , let $u$ be the inside of the bracket: $u = x^2 + 1$ . Its derivative ( $2x$ ) will cancel out the $x$ on the outside.
Exponentials: If you have $\frac{e^x}{1 + e^x}$ , let $u = 1 + e^x$ . The derivative is $e^x$ , which cancels the numerator.
Logarithms: If you have $\frac{\ln x}{x}$ , let $u = \ln x$ . The derivative is $\frac{1}{x}$ , which cancels the denominator.
Trigonometry: If you have $\sin^3 x \cos x$ , let $u = \sin x$ . The derivative is $\cos x$ , which cancels the remaining trig term.

Simple Worked Example

Question: Evaluate the definite integral $\int_0^2 \frac{x}{\sqrt{x^2 + 5}} \, dx$ .

Solution: Step 1: Choose $u$ . Let $u = x^2 + 5$ . (This is the expression inside the square root).

Step 2: Find $du$ and $dx$ . $\frac{du}{dx} = 2x$ $dx = \frac{du}{2x}$

Step 3: Change the bounds. When $x = 0$ , $u = (0)^2 + 5 = 5$ . (New lower bound). When $x = 2$ , $u = (2)^2 + 5 = 9$ . (New upper bound).

Step 4: Substitute everything into the integral. $\int_0^2 \frac{x}{\sqrt{x^2 + 5}} \, dx = \int_5^9 \frac{x}{\sqrt{u}} \cdot \frac{du}{2x}$

Step 5: Simplify and Integrate. The $x$ 's cancel out perfectly. $= \int_5^9 \frac{1}{2\sqrt{u}} \, du$ $= \frac{1}{2} \int_5^9 u^{-\frac{1}{2}} \, du$

Integrate by adding 1 to the power and dividing by the new power: $= \frac{1}{2} \left[ \frac{u^{\frac{1}{2}}}{1/2} \right]_5^9$ $= \left[ \sqrt{u} \right]_5^9$

Step 6: Evaluate using the $u$ -bounds. $= \sqrt{9} - \sqrt{5}$ $= 3 - \sqrt{5}$

Final Answer: $3 - \sqrt{5}$ .

mini-FAQ page

Q: What if the $x$ 's don't cancel out completely? A: If you have an $x$ leftover, you might need to use "algebraic substitution". Go back to your original $u = f(x)$ equation and rearrange it to make $x$ the subject. Then substitute this expression for the leftover $x$ . If this still creates a mess, your initial choice of $u$ was probably wrong.

Q: Do I have to change the bounds? Can I just integrate, back-substitute to $x$ , and use the original bounds? A: Technically yes, but it is highly discouraged. It takes much longer, creates massive algebraic expressions that are hard to evaluate, and markers are specifically looking for your ability to change limits. Always change the bounds for definite integrals.

Common mistakes to avoid

Forgetting to write $du$ : Do not write $\int \frac{1}{u} dx$ . This is mathematically meaningless because the variables are mixed. You must find $dx = \frac{du}{f'(x)}$ and substitute it so the entire integral is with respect to $u$ .
Mixing $x$ and $u$ in the bounds: Writing $\int_0^2 \frac{1}{\sqrt{u}} du$ is wrong. The numbers 0 and 2 belong to the $x$ -world, not the $u$ -world. If you write this line in an exam, you will lose a mark for incorrect notation, even if you fix it later.

Practice on Vu's Maths Hub

Fluency in substitution is the key to conquering Extension 2 calculus.

Practice identifying tricky substitutions without being prompted in the HSC Integration Booklet.
Test your skills against difficult algebraic manipulations at Vu's Maths Hub.

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