Integration by Substitution
You know how to differentiate composite functions using the chain rule. Substitution is the reverse operation — it unmasks a hidden structure so that a complicated integral collapses into a simple one. If $u = g(x)$, then $\displaystyle\int f(g(x))\,g'(x)\,dx = \int f(u)\,du$. Master this and every integral in the module becomes solvable.
Consider $\displaystyle\int 2x\cos(x^2)\,dx$. Without looking anything up — what pattern do you notice? Write down what the integrand reminds you of.
Integration by substitution reverses the chain rule. Recall that $\dfrac{d}{dx}[F(g(x))] = F'(g(x))\cdot g'(x)$. Running this backwards gives the substitution rule:
Let $u = g(x)$. Then $\dfrac{du}{dx} = g'(x)$, so we write $du = g'(x)\,dx$. The integral becomes:
$\displaystyle\int f(g(x))\,g'(x)\,dx \;=\; \int f(u)\,du$
Integrate with respect to $u$, then substitute $g(x)$ back for $u$.
Key facts
- $\displaystyle\int f(g(x))\,g'(x)\,dx = \int f(u)\,du$ where $u = g(x)$
- $du = g'(x)\,dx$ is the differential substitution step
- For definite integrals: change limits, do not back-substitute
Concepts
- Why substitution is the reverse of the chain rule
- How to identify when an integrand has the substitution structure
- The role of $g'(x)$ as the derivative "trigger" in the integrand
Skills
- Apply substitution to evaluate indefinite integrals
- Choose an appropriate $u$ from the structure of the integrand
- Recognise and handle a missing constant factor (adjust $du$)
An integral has the substitution structure when the integrand contains a composite function $f(g(x))$ multiplied by (a constant multiple of) the derivative $g'(x)$.
Example — back to the hook: Find $\displaystyle\int 2x\cos(x^2)\,dx$.
- The inner function is $x^2$. Let $u = x^2$.
- Then $du = 2x\,dx$ — and $2x\,dx$ is exactly present in the integrand!
- Substitute: $\displaystyle\int \cos(u)\,du = \sin(u) + C = \sin(x^2) + C$.
Notice: once $u$ is chosen, the $g'(x)$ factor ($2x$) absorbs into $du$ and the integral simplifies immediately.
An integral has the substitution structure when the integrand contains a composite function $f(g(x))$ multiplied by (a constant multiple of) the derivative $g'(x)$.
Pause — copy the substitution recognition criterion: look for $f(g(x))\cdot g'(x)$ and identify $u=g(x)$ so that $du=g'(x)\,dx$ into your book.
Quick check: For $\displaystyle\int 3x^2\sin(x^3)\,dx$, which substitution removes all $x$ terms?
We just saw that the substitution structure appears when the integrand is $f(g(x))\cdot g'(x)$: you can set $u=g(x)$, so $du=g'(x)\,dx$, and the integral becomes $\int f(u)\,du$. That raises a question: once you've spotted the pattern, what are the four steps to execute the substitution cleanly, including converting limits for definite integrals? This card answers it → (1) define $u$; (2) find $du$; (3) rewrite the integral; (4) integrate and back-substitute (or convert limits).
Every substitution integral follows the same four steps. Internalise them and the technique becomes automatic.
- Choose $u$. Identify the inner function or the expression whose derivative is also present.
- Differentiate. Write $du = g'(x)\,dx$ and rearrange for $dx$ if needed: $dx = \dfrac{du}{g'(x)}$.
- Substitute. Replace every $x$ expression with $u$ (including $dx \to du$). The result must be a pure $u$-integral.
- Integrate and back-substitute. Integrate with respect to $u$, then replace $u$ with $g(x)$, and add $+C$.
Example: $\displaystyle\int (2x+1)^5\,dx$.
- Let $u = 2x+1$, so $du = 2\,dx$, i.e. $dx = \tfrac{1}{2}\,du$.
- $\displaystyle\int u^5 \cdot \tfrac{1}{2}\,du = \tfrac{1}{2} \cdot \tfrac{u^6}{6} + C = \dfrac{(2x+1)^6}{12} + C$.
Every substitution integral follows the same four steps. Internalise them and the technique becomes automatic.
Pause — copy the four-step substitution method: (1) $u=g(x)$; (2) $du=g'(x)dx$; (3) rewrite as $\int f(u)\,du$; (4) integrate and back-substitute into your book.
Did you get this? True or false: $\displaystyle\int (2x+1)^5\,dx = \dfrac{(2x+1)^6}{12} + C$.
Worked examples · 3 in a row, reveal as you go
Find $\displaystyle\int (3x-2)^4\,dx$.
Find $\displaystyle\int x\sqrt{x^2+4}\,dx$.
Find $\displaystyle\int xe^{x^2}\,dx$.
Fill the gap: For $\displaystyle\int 6x^2(x^3+1)^4\,dx$, let $u = x^3+1$. Then $du = $ , so $6x^2\,dx = 2\,du$.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: when applying substitution to a definite integral $\displaystyle\int_1^2 2x(x^2+1)^3\,dx$, the correct new limits (after $u = x^2+1$) are $u=2$ and $u=5$.
Activities · practice with the ideas
Find $\displaystyle\int (5x+3)^3\,dx$ using substitution. Show all four steps clearly.
Find $\displaystyle\int \cos(3x)\,dx$. (Hint: let $u = 3x$.)
Find $\displaystyle\int 4x^3(x^4-1)^6\,dx$. Identify the substitution, showing $du$ and the simplified integral.
Find $\displaystyle\int \dfrac{x}{\sqrt{1-x^2}}\,dx$. (Hint: let $u = 1 - x^2$.)
Verify that $\dfrac{d}{dx}\!\left[\sin(x^2)\right] = 2x\cos(x^2)$ and hence write down $\displaystyle\int 2x\cos(x^2)\,dx$.
Odd one out: Three of these are correct answers. Which one is NOT?
Earlier you considered $\displaystyle\int 2x\cos(x^2)\,dx$ without any tools.
The exact answer is $\sin(x^2) + C$. The key insight: the integrand contains $\cos(x^2)$ — a composite — multiplied by $2x$, which is precisely the derivative of $x^2$. Setting $u = x^2$ absorbs both pieces at once, turning the integral into the elementary $\int\cos(u)\,du = \sin(u) + C$.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Use substitution to find $\displaystyle\int (4x+1)^3\,dx$. (2 marks)
Q2. Use substitution to find $\displaystyle\int x^2\sqrt{x^3+8}\,dx$. (3 marks)
Q3. Show that $\displaystyle\int_0^1 xe^{x^2}\,dx = \dfrac{e-1}{2}$. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $u=5x+3$, $dx=du/5$: $\tfrac{1}{5}\cdot\tfrac{u^4}{4}+C = \dfrac{(5x+3)^4}{20}+C$
2. $u=3x$, $dx=du/3$: $\tfrac{1}{3}\sin(3x)+C$
3. $u=x^4-1$, $du=4x^3\,dx$: $\displaystyle\int u^6\,du = \dfrac{u^7}{7}+C = \dfrac{(x^4-1)^7}{7}+C$
4. $u=1-x^2$, $x\,dx=-du/2$: $-\sqrt{u}+C = -\sqrt{1-x^2}+C$
5. $\dfrac{d}{dx}[\sin(x^2)] = \cos(x^2)\cdot 2x = 2x\cos(x^2)$; hence $\int 2x\cos(x^2)\,dx = \sin(x^2)+C$.
Q1 (2 marks): Let $u=4x+1$, $du=4\,dx$, $dx=du/4$ [1]. $\displaystyle\int u^3 \cdot \tfrac{1}{4}\,du = \tfrac{u^4}{16}+C = \dfrac{(4x+1)^4}{16}+C$ [1].
Q2 (3 marks): Let $u=x^3+8$, $du=3x^2\,dx$, $x^2\,dx=du/3$ [1]. $\displaystyle\int u^{1/2}\cdot\tfrac{1}{3}\,du = \tfrac{1}{3}\cdot\tfrac{2u^{3/2}}{3}+C = \dfrac{2(x^3+8)^{3/2}}{9}+C$ [2].
Q3 (3 marks): Let $u=x^2$, $du=2x\,dx$. When $x=0$, $u=0$; when $x=1$, $u=1$ [1]. $\displaystyle\int_0^1 \tfrac{1}{2}e^u\,du = \left[\tfrac{1}{2}e^u\right]_0^1 = \tfrac{1}{2}e - \tfrac{1}{2} = \dfrac{e-1}{2}$ [2].
Five timed questions on integration by substitution. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering substitution questions. Lighter alternative to the boss.
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