Substitution: Worked Examples
You know the four-step method. Now you build speed through volume — indefinite integrals, definite integrals, trig, exponentials, and the missing-constant trick. See each type done correctly once, then do it yourself. The goal: by the end of this lesson a substitution integral should feel automatic.
Consider $\displaystyle\int_0^1 3x^2(x^3+1)^5\,dx$. Before computing — estimate whether the result is bigger or smaller than 1. Write your reasoning.
When applying substitution $u = g(x)$ to a definite integral, the limits must change to match the new variable:
If $u = g(x)$ and the original limits are $x = a$ and $x = b$, the new limits become $u = g(a)$ and $u = g(b)$:
$\displaystyle\int_a^b f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du$
After evaluating the $u$-integral between the new limits, do not back-substitute — the definite integral is a number.
Key facts
- $\displaystyle\int_a^b f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du$
- Indefinite: back-substitute; definite: change limits only
- Always substitute back to the original variable for indefinite integrals
Concepts
- Why converting limits eliminates the need for back-substitution
- How the missing-constant trick applies to both indefinite and definite cases
- When to use substitution versus a standard integral formula
Skills
- Apply substitution to definite integrals with correct limit conversion
- Work through a variety of integrand types (polynomial, trig, exponential)
- Independently select, apply, and verify substitution for multi-step problems
Let us evaluate $\displaystyle\int_0^1 3x^2(x^3+1)^5\,dx$ — the estimate from card 01.
- Let $u = x^3+1$. Then $du = 3x^2\,dx$.
- Change limits: when $x=0$, $u = 0^3+1 = 1$; when $x=1$, $u = 1^3+1 = 2$.
- Substitute: $\displaystyle\int_1^2 u^5\,du = \left[\dfrac{u^6}{6}\right]_1^2 = \dfrac{64}{6} - \dfrac{1}{6} = \dfrac{63}{6} = \dfrac{21}{2}$.
The exact answer is $\dfrac{21}{2} = 10.5$ — larger than 1, because the integrand reaches $(1+1)^5 \cdot 3 = 96$ at $x = 1$.
$\int_0^1 3x^2(x^3+1)^5\,dx$: let $u=x^3+1$, $du=3x^2dx$, limits $1\to2$; $=\int_1^2 u^5\,du=\tfrac{21}{2}$.
Pause — copy the hook solution in full: $u=x^3+1$, convert limits to $u=1,2$, evaluate $\int_1^2 u^5\,du=\tfrac{63}{6}=\tfrac{21}{2}$ into your book.
Quick check: For $\displaystyle\int_0^2 2x(x^2+3)^4\,dx$ with $u = x^2+3$, the new upper limit is:
We just saw that $\int_0^1 3x^2(x^3+1)^5\,dx$: let $u=x^3+1$, $du=3x^2\,dx$; limits become $u=1$ to $u=2$; integral $=\int_1^2 u^5\,du=[u^6/6]_1^2=64/6-1/6=63/6=21/2$. That raises a question: the same technique works on trig and exponential integrands — which specific standard results do you need to complete these? This card answers it → $\int e^{u}\,du=e^u+C$; $\int\sin u\,du=-\cos u+C$; $\int\cos u\,du=\sin u+C$; $\int\sec^2 u\,du=\tan u+C$.
Substitution applies equally to trigonometric and exponential integrands. The key standard results to know:
- $\displaystyle\int \sin(u)\,du = -\cos(u) + C$
- $\displaystyle\int \cos(u)\,du = \sin(u) + C$
- $\displaystyle\int e^u\,du = e^u + C$
- $\displaystyle\int \dfrac{1}{u}\,du = \ln|u| + C$ (for $u \neq 0$)
Example — trig: $\displaystyle\int \sin^4 x \cos x\,dx$.
- Notice $\cos x = \dfrac{d}{dx}[\sin x]$. Let $u = \sin x$, $du = \cos x\,dx$.
- $\displaystyle\int u^4\,du = \dfrac{u^5}{5} + C = \dfrac{\sin^5 x}{5} + C$.
Example — exponential: $\displaystyle\int \dfrac{e^{\sqrt{x}}}{\sqrt{x}}\,dx$.
- Let $u = \sqrt{x} = x^{1/2}$. Then $du = \dfrac{1}{2\sqrt{x}}\,dx$, so $\dfrac{dx}{\sqrt{x}} = 2\,du$.
- $\displaystyle\int e^u \cdot 2\,du = 2e^u + C = 2e^{\sqrt{x}} + C$.
Substitution applies equally to trigonometric and exponential integrands. The key standard results to know:
Pause — copy the four standard substitution results for trig/exponential: $\int e^u du$, $\int\sin u\,du$, $\int\cos u\,du$, $\int\sec^2 u\,du$ into your book.
Did you get this? True or false: $\displaystyle\int \sin^3 x \cos x\,dx = \dfrac{\sin^4 x}{4} + C$.
Worked examples · 3 in a row, reveal as you go
Evaluate $\displaystyle\int_0^{\pi/2} \cos x \cdot e^{\sin x}\,dx$.
Limits: $x=0 \Rightarrow u=0$; $x=\tfrac{\pi}{2} \Rightarrow u=1$.
Find $\displaystyle\int \dfrac{x}{x^2+5}\,dx$.
Evaluate $\displaystyle\int_1^0 2x(x^2+1)^3\,dx$ (note the reversed limits).
Limits: $x=1 \Rightarrow u=2$; $x=0 \Rightarrow u=1$.
New integral: $\displaystyle\int_2^1 u^3\,du$.
Fill the gap: For $\displaystyle\int_0^2 2x(x^2+3)^4\,dx$ with $u = x^2+3$, the new upper limit is $u = $ and the new lower limit is $u = $ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: for $\displaystyle\int_0^1 3x^2(x^3+1)^5\,dx$ with $u = x^3+1$, the new limits are $u=1$ (lower) and $u=2$ (upper).
Activities · practice with the ideas
Evaluate $\displaystyle\int_0^1 2x(x^2+1)^3\,dx$ using substitution with limit conversion. Show all steps.
Find $\displaystyle\int \cos^5 x \sin x\,dx$. (Hint: let $u = \cos x$.)
Find $\displaystyle\int \dfrac{2x+1}{x^2+x+3}\,dx$. Identify the logarithmic pattern and state the answer.
Evaluate $\displaystyle\int_1^e \dfrac{\ln x}{x}\,dx$. (Hint: let $u = \ln x$, note $\dfrac{du}{dx} = \dfrac{1}{x}$.)
Find $\displaystyle\int \dfrac{e^x}{e^x+1}\,dx$. Write the substitution, $du$, and the final answer.
Odd one out: Three of these definite integral evaluations are correct. Which one is NOT?
Earlier you estimated whether $\displaystyle\int_0^1 3x^2(x^3+1)^5\,dx$ was bigger or smaller than 1.
The exact answer is $\dfrac{21}{2} = 10.5$. The integrand $(x^3+1)^5$ grows rapidly — at $x=1$ it equals $2^5 = 32$, which is multiplied by $3x^2 = 3$ to give 96. The "average height" over $[0,1]$ is roughly 10.5, explaining why the area well exceeds 1. Intuition built from boundary values helps sense-check definite integrals before you compute.
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. Evaluate $\displaystyle\int_0^1 3x^2 e^{x^3}\,dx$. (2 marks)
Q2. Find $\displaystyle\int \dfrac{3x^2}{x^3+4}\,dx$. (3 marks)
Q3. Evaluate $\displaystyle\int_0^{\pi/6} \sin(3x)\,dx$ using the substitution $u = 3x$. Show the limit conversion clearly. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $u=x^2+1$, limits $1\to 2$: $\left[\dfrac{u^4}{4}\right]_1^2 = \dfrac{16-1}{4} = \dfrac{15}{4}$
2. $u=\cos x$, $du=-\sin x\,dx$: $\displaystyle\int -u^5\,du = -\dfrac{u^6}{6}+C = -\dfrac{\cos^6 x}{6}+C$
3. Numerator $2x+1 = \dfrac{d}{dx}(x^2+x+3)$: $\ln(x^2+x+3)+C$
4. $u=\ln x$, $du=dx/x$, limits $0\to 1$: $\left[\dfrac{u^2}{2}\right]_0^1 = \dfrac{1}{2}$
5. $u=e^x+1$, $du=e^x\,dx$: $\ln(e^x+1)+C$
Odd one out — B: $\displaystyle\int_0^{\pi/2}\sin x\cos x\,dx$: with $u=\sin x$, limits $0\to 1$: $\left[\dfrac{u^2}{2}\right]_0^1 = \dfrac{1}{2}$ (not 1). So B is incorrect.
Q1 (2 marks): Let $u=x^3$, $du=3x^2\,dx$. Limits: $x=0\Rightarrow u=0$; $x=1\Rightarrow u=1$ [1]. $\displaystyle\int_0^1 e^u\,du = [e^u]_0^1 = e-1$ [1].
Q2 (3 marks): Recognise $g'/g$ pattern: numerator $3x^2 = \tfrac{d}{dx}(x^3+4)$ [1]. Let $u=x^3+4$, $du=3x^2\,dx$ [1]. $\displaystyle\int \dfrac{1}{u}\,du = \ln|u|+C = \ln(x^3+4)+C$ [1].
Q3 (3 marks): Let $u=3x$, $du=3\,dx$, $dx=du/3$ [1]. Limits: $x=0\Rightarrow u=0$; $x=\tfrac{\pi}{6}\Rightarrow u=\tfrac{\pi}{2}$ [1]. $\displaystyle\int_0^{\pi/2}\sin u\cdot\dfrac{1}{3}\,du = \dfrac{1}{3}[-\cos u]_0^{\pi/2} = \dfrac{1}{3}(0-(-1)) = \dfrac{1}{3}$ [1].
Five timed questions on definite and indefinite substitution integrals. 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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