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hscscience Ext 1 · Y12
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Module 8 · L3 of 20 ~35 min ⚡ +90 XP available

Choosing the Substitution

You've seen how substitution works — now the real skill is picking the right $u$. A well-chosen substitution turns a hard integral into a one-liner; a poorly-chosen one creates a dead end. This lesson builds your pattern-recognition: look for a composite function and its derivative lurking inside the integrand.

Today's hook — Before reading on, consider $\displaystyle\int 2x\sqrt{x^2+1}\,dx$. What would you set as $u$? Write your instinct now. You'll confirm or correct it after card 05.
0/5QUESTS
01
Recall — your gut answer first
+5 XP warm-up

Consider $\displaystyle\int 2x\sqrt{x^2+1}\,dx$. Without calculating — what expression would you choose for $u$? Write your reasoning below.

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02
The two-step recognition method
+5 XP to read

Every integration-by-substitution problem has two ingredients hiding in the integrand: a function $f(x)$ and a scalar multiple of its derivative $f'(x)$. Your job is to spot them.

The chain-rule reversal pattern:

If integrand contains $f'(x)\cdot g(f(x))$, set $u = f(x)$, so $du = f'(x)\,dx$.

The $f'(x)\,dx$ part absorbs into $du$ and the integral becomes $\displaystyle\int g(u)\,du$ — often trivial.

f ′(x) dx → du g(f(x)) → g(u) ∫ g(u) du ← simpler!
$u = f(x) \;\Rightarrow\; du = f'(x)\,dx$
Inner function rule
If you see a composite $g(f(x))$, try $u = f(x)$. The inner function is the natural choice.
Derivative check
After choosing $u$, compute $du/dx$ and verify its value (or a multiple) is present elsewhere in the integrand.
Scalar multiples are fine
If $du$ differs by a constant factor, adjust with a constant coefficient outside the integral. Constants never prevent a substitution from working.
03
What you'll master
Know

Key facts

  • The pattern $\int f'(x)[f(x)]^n\,dx = \dfrac{[f(x)]^{n+1}}{n+1} + C$
  • When to use the inner function as $u$ vs. a less obvious substitution
  • How to handle a missing constant factor in $du$
Understand

Concepts

  • Why the chain rule underpins every substitution
  • How "spotting $f'(x)$ beside $f(x)$" reduces guesswork
  • The role of changing limits for definite integrals after substitution
Can do

Skills

  • Identify a suitable substitution for integrals involving radicals and compositions
  • Verify a chosen substitution by computing $du$ and checking it appears in the integrand
  • Complete the substitution, integrate, and back-substitute correctly
04
Key terms
Substitution ($u$-sub)A technique that replaces a complex expression with a single variable $u$ to simplify an integral, reversing the chain rule.
Inner functionIn a composite $g(f(x))$, the function $f(x)$ is the inner function. It is usually the first candidate for $u$.
$du = f'(x)\,dx$The differential of $u$. Rewriting $dx$ in terms of $du$ is the key step that transforms the integral.
Back-substitutionAfter integrating in $u$, replace $u$ with $f(x)$ to express the answer in terms of the original variable.
Scalar adjustmentIf $f'(x)$ is off by a constant factor $k$, multiply and divide by $k$ so the integrand matches $du$ exactly.
Limits of integrationFor definite integrals, the limits must be converted from $x$-values to $u$-values when substituting.
05
The key recognition pattern
core concept

The substitution rule works whenever the integrand can be written in the form $f'(x) \cdot g(f(x))$. Setting $u = f(x)$ gives $du = f'(x)\,dx$, so:

$$\int f'(x)\cdot g(f(x))\,dx = \int g(u)\,du$$

The most common application is the power pattern:

$$\int f'(x)\,[f(x)]^n\,dx = \frac{[f(x)]^{n+1}}{n+1} + C \qquad (n \neq -1)$$

Worked application: $\displaystyle\int 2x\sqrt{x^2+1}\,dx$

Set $u = x^2+1$, so $du = 2x\,dx$. The factor $2x\,dx$ is already present:

$= \displaystyle\int u^{1/2}\,du = \dfrac{2}{3}u^{3/2} + C = \dfrac{2}{3}(x^2+1)^{3/2} + C$

Answer your hook question. $u = x^2 + 1$ was the right choice because its derivative $2x$ is the coefficient sitting in front of the radical. Whenever you see $(\text{something})$ under a radical and $(\text{something})'$ elsewhere in the integrand, that's your $u$.

The substitution rule works whenever the integrand can be written in the form $f'(x) \cdot g(f(x))$. Setting $u = f(x)$ gives $du = f'(x)\,dx$, so:

Pause — copy the recognition pattern: identify inner function $f(x)$, check $f'(x)$ is present as a factor, then set $u=f(x)$ into your book.

Quick check: For $\displaystyle\int 3x^2(x^3+5)^4\,dx$, the best choice of $u$ is:

06
Adjusting for a missing constant factor
core concept

We just saw the recognition pattern: look for $f'(x)\cdot g(f(x))$ in the integrand — identifying the inner function $f(x)$ and confirming its derivative $f'(x)$ is present (up to a constant). That raises a question: when $f'(x)$ is present but multiplied by the wrong constant (e.g.\ $2x$ instead of $x$ inside $x^2$), how do you balance the integral without changing its value? This card answers it → multiply inside by the needed constant and outside by its reciprocal.

Sometimes $f'(x)$ is present but multiplied by the wrong constant. You can always balance this by introducing a compensating factor outside the integral.

Example: $\displaystyle\int x(x^2+3)^5\,dx$

Set $u = x^2+3$, so $du = 2x\,dx$. We only have $x\,dx$, not $2x\,dx$. Rewrite:

$x\,dx = \tfrac{1}{2}\,du$

$$\int x(x^2+3)^5\,dx = \frac{1}{2}\int u^5\,du = \frac{u^6}{12} + C = \frac{(x^2+3)^6}{12} + C$$

The $\tfrac{1}{2}$ factor is introduced outside the integral to compensate for the missing 2 in $du$.

The golden rule. You may multiply and divide by any non-zero constant to match $du$ — but you may never introduce or remove a factor involving $x$ in this way. Constants only.

Sometimes $f'(x)$ is present but multiplied by the wrong constant. You can always balance this by introducing a compensating factor outside the integral.

Pause — copy the constant-adjustment technique: $\int 2x\cdot f(x^2)\,dx$ vs $\int x\cdot f(x^2)\,dx$ — introduce $\frac{1}{2}\cdot 2$ to match the derivative into your book.

Did you get this? True or false: for $\displaystyle\int x^2(x^3-1)^7\,dx$, setting $u = x^3-1$ works because $du = 3x^2\,dx$ and $x^2\,dx = \tfrac{1}{3}du$.

PROBLEM 1 · POWER PATTERN

Find $\displaystyle\int 6x^2(2x^3-1)^4\,dx$.

1
Spot the inner function: $u = 2x^3 - 1$, so $du = 6x^2\,dx$.
$6x^2$ is exactly the derivative of $2x^3-1$, so this is a perfect match — no adjustment needed.
PROBLEM 2 · RADICAL WITH ADJUSTMENT

Find $\displaystyle\int \frac{x}{\sqrt{x^2+4}}\,dx$.

1
Set $u = x^2+4$, so $du = 2x\,dx$, giving $x\,dx = \tfrac{1}{2}du$.
The inner function under the radical is $x^2+4$; its derivative $2x$ differs from $x$ by a factor of 2.
PROBLEM 3 · DEFINITE INTEGRAL

Evaluate $\displaystyle\int_0^2 x(x^2+1)^3\,dx$.

1
Set $u = x^2+1$, $du = 2x\,dx$, so $x\,dx = \tfrac{1}{2}du$. Change limits: $x=0 \Rightarrow u=1$; $x=2 \Rightarrow u=5$.
For a definite integral, always convert the limits to $u$-values. This avoids back-substitution at the end.

Fill the gap: For $\displaystyle\int x(x^2-3)^2\,dx$, setting $u = x^2-3$ gives $du = 2x\,dx$, so the integral becomes $\dfrac{1}{2}\displaystyle\int u^2\,du = \dfrac{u^3}{$$} + C$.

Trap 01
Forgetting to substitute $dx$
Substitution transforms the entire integrand, including the $dx$. Students often replace $f(x)$ with $u$ but forget to replace $dx$ with $\tfrac{du}{f'(x)}$. The $dx$ must go — leaving it produces a mixed-variable integral that cannot be evaluated.
Trap 02
Introducing a variable factor to "fix" $du$
If $du = 2x\,dx$ but you only have $dx$ (not $x\,dx$), you cannot multiply inside by $x$ and divide outside by $x$ — that changes the integrand. You can only compensate with constants. If a variable factor is missing, try a different substitution.
Trap 03
Forgetting to change limits
For definite integrals, the limits $a$ and $b$ are $x$-values. After substituting $u = f(x)$, the new limits must be $f(a)$ and $f(b)$. Failing to change limits and then evaluating in $u$ gives the wrong answer.

Did you get this? True or false: when evaluating $\displaystyle\int_1^3 2x(x^2+1)^3\,dx$ via $u = x^2+1$, the new limits are $u = 2$ and $u = 10$.

Work mode · how are you completing this lesson?
1

State a suitable substitution $u$ for $\displaystyle\int \cos x \cdot \sin^3 x\,dx$ and explain why it works.

2

Find $\displaystyle\int 4x(x^2-7)^3\,dx$.

3

Find $\displaystyle\int \frac{x^2}{\sqrt{x^3+1}}\,dx$.

4

Evaluate $\displaystyle\int_0^1 3x^2(x^3+2)^4\,dx$. Change the limits.

5

Explain why $u = x$ is never a useful substitution. What does a useful $u$ always achieve?

Odd one out: Three of these substitutions will work for their respective integrals. Which one will NOT?

11
Revisit your thinking

Earlier you chose a substitution for $\displaystyle\int 2x\sqrt{x^2+1}\,dx$.

The correct choice is $u = x^2+1$, because its derivative $2x$ is the coefficient of the radical in the integrand. This gives $\displaystyle\int \sqrt{u}\,du = \tfrac{2}{3}u^{3/2} + C = \tfrac{2}{3}(x^2+1)^{3/2} + C$. Did you spot the pattern?

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01
Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

02
Short answer
ApplyBand 31 mark

Q1. State a suitable substitution for $\displaystyle\int 5x^4(x^5+3)^2\,dx$ and find the integral. (1 mark)

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ApplyBand 42 marks

Q2. Find $\displaystyle\int \frac{x}{\sqrt{x^2-9}}\,dx$, $x > 3$. Show your substitution and adjustment. (2 marks)

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AnalyseBand 52 marks

Q3. Evaluate $\displaystyle\int_0^{\pi/2} \sin^3 x \cos x\,dx$. Change the limits using your substitution. (2 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $u = \sin x$, $du = \cos x\,dx$ — perfect match. 2. $u = x^2-7$, $du = 2x\,dx$; $\int 4x(x^2-7)^3\,dx = 2\int u^3\,du = \frac{u^4}{2}+C = \frac{(x^2-7)^4}{2}+C$. 3. $u = x^3+1$, $du = 3x^2\,dx$; $x^2\,dx = \frac{1}{3}du$; $\frac{1}{3}\int u^{-1/2}\,du = \frac{2}{3}u^{1/2}+C = \frac{2}{3}\sqrt{x^3+1}+C$. 4. $u = x^3+2$; limits $u=2$ to $u=3$; $\left[\frac{u^5}{5}\right]_2^3 = \frac{243-32}{5} = \frac{211}{5}$. 5. $u = x$ gives $du = dx$ so nothing simplifies; a useful $u$ must simplify the integrand by replacing a composite expression.

Q1 (1 mark): $u = x^5+3$, $du = 5x^4\,dx$; $\int u^2\,du = \dfrac{u^3}{3}+C = \dfrac{(x^5+3)^3}{3}+C$ [1].

Q2 (2 marks): $u = x^2-9$, $du = 2x\,dx$, $x\,dx = \frac{1}{2}du$ [1]; $\frac{1}{2}\int u^{-1/2}\,du = u^{1/2}+C = \sqrt{x^2-9}+C$ [1].

Q3 (2 marks): $u = \sin x$, $du = \cos x\,dx$; limits: $x=0\Rightarrow u=0$, $x=\frac{\pi}{2}\Rightarrow u=1$ [1]; $\int_0^1 u^3\,du = \left[\frac{u^4}{4}\right]_0^1 = \frac{1}{4}$ [1].

01
Boss battle · The Substitution Strategist
earn bronze · silver · gold

Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena
02
Science Jump · platform challenge

Climb platforms by answering substitution questions. A lighter alternative to the boss battle.

Mark lesson as complete

Tick when you've finished the practice and review.

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