Integration by Substitution — Review and Extensions
Substitution is the chain rule run backwards. The instinct to look for: a piece of the integrand whose derivative also appears (up to a constant). When you can spot that pattern, $\int f(g(x))\, g'(x)\, dx$ collapses to $\int f(u)\, du$. This lesson reviews the mechanics of $u$-substitution, then extends to patterns where the simplification is less obvious — including reverse-engineering $du$ to make the algebra fit.
Differentiate $(x^2+1)^6$ using the chain rule. Before checking — what does this tell you about $\int 12x(x^2+1)^5\,dx$? Write the derivative and the antiderivative below.
Substitution rewards two habits: spot the inner function whose derivative also lives inside the integrand, then replace everything in sight — including the $dx$ — using $du = g'(x)\,dx$. If anything in $x$ remains after substituting, you've chosen the wrong $u$.
The spot-replace-rearrange reading: (1) identify $u = g(x)$ where $g'(x)$ also appears (possibly off by a constant), (2) compute $du = g'(x)\,dx$ and solve for $dx$ if needed, (3) rewrite the integral purely in $u$ before integrating.
$\int f(g(x))\,g'(x)\,dx = \int f(u)\,du$ · $u = g(x)$ · $du = g'(x)\,dx$
Key facts
- Substitution rule: $\int f(g(x))\,g'(x)\,dx = \int f(u)\,du$ with $u = g(x)$
- $du = g'(x)\,dx$ is treated as a differential identity
- For definite integrals, limits transform: $x = a \to u = g(a)$, $x = b \to u = g(b)$
- Multiplicative constants can be moved outside the integral freely
Concepts
- Why substitution is just the chain rule running in reverse
- How to recognise an $f(g(x))\,g'(x)$ pattern when the derivative is hidden by a constant
- When substitution doesn't simplify cleanly — and what to do (algebra in $u$, or a different $u$)
Skills
- Carry out $u$-substitution with $dx$ replaced correctly using $du$
- Apply standard patterns: $\int x\sqrt{x^2 \pm a^2}\,dx$, $\int \sin^n x \cos x\,dx$, $\int \tfrac{f'(x)}{f(x)}\,dx$
- Change limits cleanly for definite integrals
The chain rule says $\dfrac{d}{dx}\bigl[F(g(x))\bigr] = F'(g(x))\,g'(x)$. Integrating both sides gives the substitution rule:
The mechanical procedure:
- Choose $u = g(x)$ — usually the inner function of a composite.
- Differentiate: $du = g'(x)\,dx$, then solve for $dx$ if needed: $dx = \dfrac{du}{g'(x)}$.
- Rewrite the integral in $u$ — every $x$ must vanish.
- Integrate in $u$, then back-substitute $u = g(x)$.
Worked through the hook: For $\int 2x(x^2+1)^5\,dx$, let $u = x^2 + 1$. Then $du = 2x\,dx$, so $2x\,dx$ in the integrand is literally $du$. The integral becomes $\int u^5\,du = \tfrac{u^6}{6} + C = \tfrac{(x^2+1)^6}{6} + C$.
Substitution rule: $\int f(g(x))\,g'(x)\,dx = \int f(u)\,du$, $u = g(x)$ · Procedure: choose $u$, find $du$, rewrite in $u$, integrate, back-substitute · $\int 2x(x^2+1)^5\,dx = \tfrac{(x^2+1)^6}{6} + C$ · A stray $x$ after substituting means the wrong $u$ — or expand the bracket instead
Pause — copy the substitution rule $\int f(g(x))\,g'(x)\,dx = \int f(u)\,du$, the four-step procedure, and the worked example $\int 2x(x^2+1)^5\,dx = (x^2+1)^6/6+C$ into your book.
Quick check: For $\int 3x^2 \cos(x^3)\,dx$, what is the best choice of $u$?
We just saw the substitution rule $\int f(g(x))\,g'(x)\,dx = \int f(u)\,du$ with $u = g(x)$: choose $u$, find $du$, rewrite, integrate, back-substitute. That raises a question: what do you do when, after substituting, a stray $x$ remains or the constant coefficient doesn't match? This card answers it → convert the stray $x$ via $x = g^{-1}(u)$, or pull out $1/k$ for a constant mismatch; the $f'/f$ pattern gives $\ln|f(x)|+C$.
Not every substitution gives a one-line answer. Two common situations:
- Stray $x$ remains. When $du = g'(x)\,dx$ doesn't absorb every $x$, re-express the leftover using $u = g(x) \Rightarrow x = g^{-1}(u)$. Example: $\int x \sqrt{x+1}\,dx$ with $u = x+1$ leaves an $x = u - 1$, giving $\int (u-1)\sqrt{u}\,du = \int (u^{3/2} - u^{1/2})\,du$.
- The derivative is off by a constant. If the integrand has $g'(x)$ multiplied by some constant $k$, just pull $\tfrac{1}{k}$ outside. Example: $\int x \, e^{x^2}\,dx$ with $u = x^2$ gives $du = 2x\,dx$, so $x\,dx = \tfrac{1}{2}\,du$ and the integral becomes $\tfrac{1}{2}\int e^u\,du$.
Stray $x$ after substitution → use $x = g^{-1}(u)$ to convert · Constant mismatch → pull $\tfrac{1}{k}$ outside the integral · $\int \tfrac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C$ — the logarithmic pattern · $\int \tan x\,dx = -\ln|\cos x| + C = \ln|\sec x| + C$
Pause — copy the stray-$x$ fix ($x = g^{-1}(u)$), the constant-mismatch fix (pull out $1/k$), the logarithmic pattern $\int f'(x)/f(x)\,dx = \ln|f(x)|+C$, and $\int\tan x\,dx = -\ln|\cos x|+C$ into your book.
Did you get this? True or false: $\int \dfrac{2x+3}{x^2+3x+5}\,dx = \ln|x^2 + 3x + 5| + C$.
Worked examples · 3 in a row, reveal as you go
Evaluate $\displaystyle \int x\sqrt{x^2 + 1}\,dx$.
Evaluate $\displaystyle \int \sin^3 x \cos x\,dx$.
Evaluate $\displaystyle \int \frac{1}{x \ln x}\,dx$ for $x > 1$.
Fill the gap: For $\displaystyle \int e^{x^2}\cdot x\,dx$, the substitution $u = x^2$ gives $du = $ , so $x\,dx = $ and the integral becomes $\tfrac{1}{2}\int e^u\,du$.
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: for $\displaystyle \int_0^1 2x(x^2+1)^3\,dx$ with $u = x^2 + 1$, the new limits become $u = 1$ (when $x=0$) and $u = 2$ (when $x=1$).
Activities · practice with the ideas
Evaluate $\displaystyle \int 2x(x^2 + 3)^4\,dx$ using a single substitution. Show your $u$ and $du$.
Evaluate $\displaystyle \int \cos^4 x \sin x\,dx$.
Evaluate $\displaystyle \int \frac{x}{x^2 + 4}\,dx$ using a $\tfrac{f'}{f}$ substitution.
Evaluate $\displaystyle \int_0^{\pi/2} \sin^5 x \cos x\,dx$ by changing variable and limits.
Evaluate $\displaystyle \int x\sqrt{x + 1}\,dx$. (Stray $x$ — use $x = u - 1$.)
Odd one out: Three of these integrals yield to the substitution $u = x^2 + 1$ cleanly (no stray $x$). Which one does NOT?
Earlier you compared $\int 2x(x^2+1)^5\,dx$ with $\int (x^2+1)^5\,dx$ — guessing what $u$ would do in each.
The first yields immediately to $u = x^2+1$ because $2x\,dx = du$ is already in the integrand: $\tfrac{(x^2+1)^6}{6} + C$. The second has no $2x$ to absorb $du$ — the substitution leaves a stray $x$ that re-expresses awkwardly via $x = \sqrt{u-1}$. The reliable approach there is to expand $(x^2+1)^5$ binomially and integrate term by term. The lesson: substitution is only "easy" when the derivative of $g(x)$ is already present (up to a constant).
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 $\displaystyle \int 6x^2 (x^3 + 4)^7\,dx$ using a single substitution. (2 marks)
Q2. Evaluate $\displaystyle \int_0^{\pi/2} \sin^2 x \cos x\,dx$ by changing variable and transforming the limits. (3 marks)
Q3. Evaluate $\displaystyle \int x\sqrt{2x + 1}\,dx$. (Hint: a stray $x$ will remain after substituting — re-express it.) (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. Let $u = x^2 + 3$, $du = 2x\,dx$. $\int 2x(x^2+3)^4\,dx = \int u^4\,du = \tfrac{u^5}{5} + C = \tfrac{(x^2+3)^5}{5} + C$.
2. Let $u = \cos x$, $du = -\sin x\,dx$ so $\sin x\,dx = -du$. $\int \cos^4 x \sin x\,dx = -\int u^4\,du = -\tfrac{u^5}{5} + C = -\tfrac{\cos^5 x}{5} + C$.
3. Let $u = x^2 + 4$, $du = 2x\,dx$ so $x\,dx = \tfrac{1}{2}du$. $\int \tfrac{x}{x^2+4}\,dx = \tfrac{1}{2}\int \tfrac{1}{u}\,du = \tfrac{1}{2}\ln(x^2+4) + C$ ($x^2+4 > 0$, drop $|\,|$).
4. Let $u = \sin x$, $du = \cos x\,dx$. Limits: $x=0 \to u=0$; $x=\pi/2 \to u=1$. $\int_0^1 u^5\,du = \tfrac{1}{6}$.
5. Let $u = x + 1$, $du = dx$, $x = u - 1$. $\int x\sqrt{x+1}\,dx = \int (u-1)\sqrt{u}\,du = \int (u^{3/2} - u^{1/2})\,du = \tfrac{2}{5}u^{5/2} - \tfrac{2}{3}u^{3/2} + C = \tfrac{2}{5}(x+1)^{5/2} - \tfrac{2}{3}(x+1)^{3/2} + C$.
Q1 (2 marks): $u = x^3 + 4$, $du = 3x^2\,dx$, so $6x^2\,dx = 2\,du$ [1]. Integral $= 2\int u^7\,du = \tfrac{u^8}{4} + C = \tfrac{(x^3+4)^8}{4} + C$ [1].
Q2 (3 marks): $u = \sin x$, $du = \cos x\,dx$ [1]. Limits transform: $x=0 \to u=0$, $x=\pi/2 \to u=1$ [1]. $\int_0^1 u^2\,du = \tfrac{1}{3}$ [1].
Q3 (3 marks): $u = 2x+1$, $du = 2dx$, $x = \tfrac{u-1}{2}$ [1]. Integral $= \tfrac{1}{4}\int (u^{3/2} - u^{1/2})\,du = \tfrac{1}{4}\bigl[\tfrac{2u^{5/2}}{5} - \tfrac{2u^{3/2}}{3}\bigr] + C$ [1] $= \tfrac{(2x+1)^{5/2}}{10} - \tfrac{(2x+1)^{3/2}}{6} + C$ [1].
Five timed questions on $u$-substitution, derivative-spotting, and definite-integral limit transformation. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering quick substitution questions. Lighter alternative to the boss.
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