Quick Recall
Quick check — pick the correct statement in each question before we meet today's wall.
The chain rule says $\dfrac{\text{d}y}{\text{d}x} = \dfrac{\text{d}y}{\text{d}u} \cdot \dfrac{\text{d}u}{\text{d}x}$ — differentiate the outer function, then multiply by the derivative of the inner function.
The W6 shortcut worked because $ax+b$ is linear: its derivative $a$ is a constant, so the missing factor could be repaired by a fixed $\dfrac{1}{a}$. Today the inner function is non-linear, so a constant fix-up is no longer enough.
The Wall
Consider the integral
$$\int 2x \, e^{x^2} \, \text{d}x$$
This is a product, so last lesson's integration by parts (LIATE) might come to mind. But trying $u = 2x$, $\text{d}v = e^{x^2}\text{d}x$ fails — we cannot even find $v = \int e^{x^2} \text{d}x$! The W6 reverse-chain-rule shortcut (multiply by $\dfrac{1}{a}$) also fails because the inner function is $x^2$, not $ax+b$.
And yet — the answer is just $e^{x^2} + C$. Notice that the inner function is $x^2$, and its derivative $2x$ is sitting right next to $e^{x^2}$ as a factor. This is not a coincidence.
The Need
We need a general method for reversing the chain rule when the inner function is anything — not just $ax+b$. The method is called integration by substitution: introduce a new variable $u = g(x)$, compute $\text{d}u = g'(x)\text{d}x$, and rewrite the entire integral in terms of $u$.
💡 The Bridge (click to reveal)
The chain rule says: $\dfrac{\text{d}}{\text{d}x} F(g(x)) = F'(g(x)) \cdot g'(x)$. Integrating both sides:
$$F(g(x)) + C = \int F'(g(x)) \cdot g'(x) \, \text{d}x$$
Substituting $u = g(x)$ and $\text{d}u = g'(x)\text{d}x$, the right side becomes $\int F'(u) \, \text{d}u = F(u) + C = F(g(x)) + C$. This is the substitution rule:
$$\int f(g(x)) \, g'(x) \, \text{d}x = \int f(u) \, \text{d}u \quad \text{where } u = g(x)$$
For $\int 2x \, e^{x^2} \, \text{d}x$: let $u = x^2$, so $\text{d}u = 2x\,\text{d}x$. Then $\int 2x \, e^{x^2} \, \text{d}x = \int e^u \, \text{d}u = e^u + C = e^{x^2} + C$. The factor $2x$ was the “fingerprint” of the substitution.
Integration by substitution reverses the chain rule. Given an integral of the form $\int f(g(x)) \, g'(x) \, \text{d}x$, we introduce a new variable $u = g(x)$, compute $\text{d}u = g'(x)\text{d}x$, and rewrite the integral as $\int f(u) \, \text{d}u$ — which is usually simpler.
Let $u = g(x)$ be a differentiable function. Then $\text{d}u = g'(x) \, \text{d}x$, and:
$$\boxed{\;\int f(g(x)) \, g'(x) \, \text{d}x = \int f(u) \, \text{d}u \quad \text{where } u = g(x)\;}$$The recipe: (1) choose $u = g(x)$ (usually the inner function of a composite); (2) compute $\text{d}u = g'(x)\text{d}x$; (3) rewrite the integrand and $\text{d}x$ in terms of $u$; (4) integrate with respect to $u$.
For a definite integral $\displaystyle\int_{x=a}^{x=b} f(g(x)) \, g'(x) \, \text{d}x$, substitute $u = g(x)$ and also change the limits:
$$\int_{x=a}^{x=b} f(g(x)) \, g'(x) \, \text{d}x = \int_{u=g(a)}^{u=g(b)} f(u) \, \text{d}u$$
Lower limit $x = a \Rightarrow u = g(a)$; upper limit $x = b \Rightarrow u = g(b)$. After changing limits, you do not need to convert back to $x$.
For an indefinite integral, after integrating in $u$ you must substitute back $u = g(x)$ to express the answer in terms of the original variable $x$:
$$\int f(g(x)) \, g'(x) \, \text{d}x = \int f(u) \, \text{d}u = F(u) + C = F(g(x)) + C$$
Always end with “$+ C$” for an indefinite integral, and verify by differentiating your final answer in $x$.
- Identify the inner function $u = g(x)$ of a composite.
- Compute $\text{d}u = g'(x) \, \text{d}x$ and solve for $\text{d}x$ if needed.
- Check that $g'(x)$ appears (up to a constant factor) in the integrand.
- Rewrite the integral entirely in terms of $u$ (no $x$ should remain).
- For a definite integral: change the limits $x \to u$ and evaluate directly.
- For an indefinite integral: integrate in $u$, then substitute back $u = g(x)$ and add $+ C$.
- Always verify by differentiating your final answer.
You are locked in the substitution lab. Three sealed doors each guard one definite integral that must be cracked by a change of variable, and each answer is a whole number. The vault in the middle opens only when the group enters the sum of the three door numbers. Work in teams of three: one member owns each door, and the vault is opened together. Change the limits $x \to u$ before you evaluate.
Enter the sum of the numbers revealed by the three doors.
Team rule: the owner of each door explains the substitution and the changed limits to the group before anyone types the number. A wrong number costs nothing but time — the clock is your opponent, not a failure.
Work on your own and approve each line before it is shown. At every stage, pick the line you believe comes next in the solution of $\displaystyle\int x^{2}\sqrt{x^{3}+1}\,\text{d}x$. A correct pick reveals the line and unlocks the next stage; a wrong pick turns red and explains the trap. A near-perfect run — at most two wrong picks — earns the top band.
Solo check: when the final line is revealed, differentiate the answer once on paper and confirm that you recover the original integrand $x^{2}\sqrt{x^{3}+1}$.
Hence, or otherwise, show that $\displaystyle\int_{-1}^{3} \frac{1}{2x+3} \, \text{d}x = \ln 3$ using the substitution $u = 2x + 3$.
(Originally solved in W6 by recognising the antiderivative $\ln|2x+3|$. Here we re-solve it using a formal substitution and changing the limits.)
👁 Reveal Solution
Let $u = 2x + 3$. Differentiate: $\text{d}u = 2 \, \text{d}x$, so $\text{d}x = \dfrac{\text{d}u}{2}$.
Lower limit: $x = -1 \Rightarrow u = 2(-1) + 3 = 1$.
Upper limit: $x = 3 \Rightarrow u = 2(3) + 3 = 9$.
$\displaystyle\int_{-1}^{3} \frac{1}{2x+3} \, \text{d}x = \int_{1}^{9} \frac{1}{u} \cdot \frac{\text{d}u}{2} = \frac{1}{2}\int_{1}^{9} \frac{1}{u} \, \text{d}u$.
$\dfrac{1}{2}\displaystyle\int_{1}^{9} \frac{1}{u} \, \text{d}u = \frac{1}{2}\Big[\ln |u|\Big]_{1}^{9} = \frac{1}{2}\left(\ln 9 - \ln 1\right)$.
Since $\ln 1 = 0$ and $\ln 9 = \ln(3^2) = 2\ln 3$:
$= \dfrac{1}{2}(2\ln 3 - 0) = \ln 3$.
$\boxed{\displaystyle\int_{-1}^{3} \frac{1}{2x+3} \, \text{d}x = \ln 3}$
Verification (W6 method): antiderivative is $\frac{1}{2}\ln|2x+3|$. Evaluated: $\frac{1}{2}(\ln 9 - \ln 1) = \frac{1}{2}(2\ln 3) = \ln 3$. ✓ Numerically: $\ln 3 \approx 1.0986$.
Find the exact value of $\displaystyle\int_{0}^{\frac{\pi}{2}} \sin^2 2x \, \cos x \, \text{d}x$ using the substitution $u = \sin x$.
(Hint: first use the double-angle identity $\sin 2x = 2\sin x\cos x$ to rewrite $\sin^2 2x$, then change the limits from $x$ to $u$.)
👁 Reveal Solution
$\sin 2x = 2\sin x\cos x$, so $\sin^2 2x = (2\sin x\cos x)^2 = 4\sin^2 x\cos^2 x$.
Hence $\displaystyle\int_{0}^{\frac{\pi}{2}} \sin^2 2x\cos x \, \text{d}x = 4\int_{0}^{\frac{\pi}{2}} \sin^2 x\cos^3 x \, \text{d}x$.
Let $u = \sin x$. Then $\text{d}u = \cos x \, \text{d}x$, and $\cos^2 x = 1 - \sin^2 x = 1 - u^2$.
So $\cos^3 x \, \text{d}x = \cos^2 x \cdot \cos x \, \text{d}x = (1 - u^2) \, \text{d}u$.
Lower limit: $x = 0 \Rightarrow u = \sin 0 = 0$.
Upper limit: $x = \dfrac{\pi}{2} \Rightarrow u = \sin\dfrac{\pi}{2} = 1$.
$4\displaystyle\int_{0}^{\frac{\pi}{2}} \sin^2 x\cos^3 x \, \text{d}x = 4\int_{0}^{1} u^2(1 - u^2) \, \text{d}u = 4\int_{0}^{1} (u^2 - u^4) \, \text{d}u$.
$4\displaystyle\int_{0}^{1} (u^2 - u^4) \, \text{d}u = 4\left[\dfrac{u^3}{3} - \dfrac{u^5}{5}\right]_{0}^{1}$.
$= 4\left(\dfrac{1}{3} - \dfrac{1}{5}\right) = 4 \cdot \dfrac{2}{15} = \dfrac{8}{15}$.
$\boxed{\displaystyle\int_{0}^{\frac{\pi}{2}} \sin^2 2x\cos x \, \text{d}x = \dfrac{8}{15}}$
Verification: $\dfrac{\text{d}}{\text{d}x}\!\left(\dfrac{4}{3}\sin^3 x - \dfrac{4}{5}\sin^5 x\right) = 4\sin^2 x\cos x - 4\sin^4 x\cos x = 4\sin^2 x\cos x(1 - \sin^2 x) = 4\sin^2 x\cos^3 x = \sin^2 2x\cos x$. ✓ Numerically: $\dfrac{8}{15} \approx 0.5333$.
Attempt each question on paper first, then click Reveal Solution to check your work. Questions are ordered from easiest to hardest.
Find $\displaystyle\int e^{-4x} \, \text{d}x$ using the substitution $u = -4x$.
(Watch the coefficient: differentiating $u = -4x$ produces a negative constant, which must be compensated.)
💡 Hint
Let $u = -4x$. Then $\text{d}u = -4 \, \text{d}x$, so $\text{d}x = -\dfrac{\text{d}u}{4}$.
Substitute both $e^{-4x} = e^{u}$ and $\text{d}x$ before integrating.
👁 Reveal Solution
Let $u = -4x$. Differentiate: $\text{d}u = -4 \, \text{d}x$, so $\text{d}x = -\dfrac{\text{d}u}{4}$.
$\displaystyle\int e^{-4x} \, \text{d}x = \int e^{u} \left(-\dfrac{\text{d}u}{4}\right) = -\dfrac{1}{4}\int e^{u} \, \text{d}u$.
$-\dfrac{1}{4}\displaystyle\int e^{u} \, \text{d}u = -\dfrac{1}{4}e^{u} + C$.
$= -\dfrac{1}{4}e^{-4x} + C$.
$\boxed{-\dfrac{1}{4}e^{-4x} + C}$
Verification: $\dfrac{\text{d}}{\text{d}x}\!\left(-\dfrac{1}{4}e^{-4x}\right) = -\dfrac{1}{4}(-4)e^{-4x} = e^{-4x}$. ✓
Find $\displaystyle\int \frac{x}{x^2+1} \, \text{d}x$ using the substitution $u = x^2 + 1$.
💡 Hint
Let $u = x^2 + 1$. Then $\text{d}u = 2x \, \text{d}x$, so $x \, \text{d}x = \dfrac{\text{d}u}{2}$.
The numerator $x \, \text{d}x$ becomes $\dfrac{\text{d}u}{2}$, and the denominator $x^2 + 1$ becomes $u$.
👁 Reveal Solution
Let $u = x^2 + 1$. Then $\text{d}u = 2x \, \text{d}x$, so $x \, \text{d}x = \dfrac{\text{d}u}{2}$.
$\displaystyle\int \frac{x}{x^2+1} \, \text{d}x = \int \frac{1}{x^2+1} \cdot x \, \text{d}x = \int \frac{1}{u} \cdot \frac{\text{d}u}{2} = \frac{1}{2}\int \frac{1}{u} \, \text{d}u$.
$\dfrac{1}{2}\displaystyle\int \frac{1}{u} \, \text{d}u = \frac{1}{2}\ln|u| + C$.
Since $x^2 + 1 > 0$ for all real $x$, $|x^2 + 1| = x^2 + 1$:
$= \dfrac{1}{2}\ln(x^2 + 1) + C$.
$\boxed{\dfrac{1}{2}\ln(x^2 + 1) + C}$
Verification: $\dfrac{\text{d}}{\text{d}x}\!\left[\dfrac{1}{2}\ln(x^2+1)\right] = \dfrac{1}{2} \cdot \dfrac{2x}{x^2+1} = \dfrac{x}{x^2+1}$. ✓
Find $\displaystyle\int e^{x^2} \cdot 2x \, \text{d}x$ using the substitution $u = x^2$.
(Note: $\int e^{x^2} \, \text{d}x$ cannot be expressed in elementary functions — the factor $2x$ is essential.)
💡 Hint
Let $u = x^2$. Then $\text{d}u = 2x \, \text{d}x$ — this is exactly the “other factor” in the integrand.
The integral simplifies directly to $\int e^u \, \text{d}u$.
👁 Reveal Solution
Let $u = x^2$. Then $\text{d}u = 2x \, \text{d}x$.
$\displaystyle\int e^{x^2} \cdot 2x \, \text{d}x = \int e^u \, \text{d}u$.
(The factor $2x \, \text{d}x$ is exactly $\text{d}u$, and $e^{x^2} = e^u$.)
$\displaystyle\int e^u \, \text{d}u = e^u + C$.
$= e^{x^2} + C$.
$\boxed{e^{x^2} + C}$
Verification: $\dfrac{\text{d}}{\text{d}x}\!\left[e^{x^2}\right] = e^{x^2} \cdot 2x$. ✓ This is the “clean” case — the derivative of the inner function appears exactly as a factor in the integrand.
Find $\displaystyle\int \frac{\cos(\sqrt{x}\,)}{\sqrt{x}} \, \text{d}x$ using the substitution $u = \sqrt{x}$.
(Requires careful handling of $\text{d}u$ for a non-polynomial substitution.)
💡 Hint 1
Let $u = \sqrt{x} = x^{1/2}$. Differentiate: $\text{d}u = \dfrac{1}{2}x^{-1/2} \, \text{d}x = \dfrac{1}{2\sqrt{x}} \, \text{d}x$.
Solve for the “missing piece” in the integrand: $\dfrac{1}{\sqrt{x}} \, \text{d}x = 2 \, \text{d}u$.
💡 Hint 2
Substitute: $\cos(\sqrt{x}\,) = \cos u$ and $\dfrac{1}{\sqrt{x}} \, \text{d}x = 2 \, \text{d}u$.
The integral becomes $\int \cos u \cdot 2 \, \text{d}u = 2\int \cos u \, \text{d}u$.
👁 Reveal Solution
Let $u = \sqrt{x} = x^{1/2}$. Differentiate: $\text{d}u = \dfrac{1}{2}x^{-1/2} \, \text{d}x = \dfrac{1}{2\sqrt{x}} \, \text{d}x$.
Rearrange: $\dfrac{\text{d}x}{\sqrt{x}} = 2 \, \text{d}u$.
Split the integrand: $\dfrac{\cos(\sqrt{x}\,)}{\sqrt{x}} \, \text{d}x = \cos(\sqrt{x}\,) \cdot \dfrac{\text{d}x}{\sqrt{x}} = \cos u \cdot 2 \, \text{d}u$.
So: $\displaystyle\int \frac{\cos(\sqrt{x}\,)}{\sqrt{x}} \, \text{d}x = 2\int \cos u \, \text{d}u$.
$2\displaystyle\int \cos u \, \text{d}u = 2\sin u + C$.
$= 2\sin(\sqrt{x}\,) + C$.
$\boxed{2\sin(\sqrt{x}\,) + C}$
Verification: $\dfrac{\text{d}}{\text{d}x}\!\left[2\sin(\sqrt{x}\,)\right] = 2\cos(\sqrt{x}\,) \cdot \dfrac{1}{2\sqrt{x}} = \dfrac{\cos(\sqrt{x}\,)}{\sqrt{x}}$. ✓
Exit Ticket
Before you leave, tick the objectives you are confident you can achieve. Be honest — this guides next lesson's review.