What do you do when the integral doesn't match any standard formula? When $\int 2x \cos(x^2) \, dx$ stares back at you and the power rule, exponential rule, and logarithmic rule all fail? This is where substitution comes in — the integration equivalent of a change of variables. It's the reverse of the chain rule: just as the chain rule handles composite functions in differentiation, substitution untangles composite functions in integration. Master this technique, and a vast new territory of integrals becomes accessible.
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Consider $\int 2x \cos(x^2) \, dx$. If you let $u = x^2$, then $du = 2x \, dx$. How does this substitution simplify the integral? What would the integral become in terms of $u$?
Core Content
The substitution rule states:
∫ f(g(x)) · g'(x) dx = ∫ f(u) du where u = g(x)
Step-by-step process:
For definite integrals: Change the limits: when $x = a$, $u = g(a)$; when $x = b$, $u = g(b)$. Then evaluate directly — no need to substitute back.
Example (indefinite): $\int 2x \cos(x^2) \, dx$
Let u = x^2, then du = 2x dx
∫ 2x cos(x^2) dx = ∫ cos(u) du = sin(u) + C = sin(x^2) + C
Example (definite): $\int_0^2 x e^{x^2} \, dx$
Let u = x^2, du = 2x dx, so x dx = du/2
When x = 0, u = 0; when x = 2, u = 4
∫_0^2 x e^{x^2} dx = (1/2)∫_0^4 e^u du = (1/2)[e^u]_0^4 = (1/2)(e^4 - 1)
The simplest and most common substitution is linear: $u = ax + b$.
Example: $\int (3x + 2)^4 \, dx$
Let u = 3x + 2, then du = 3 dx, so dx = du/3
∫ (3x + 2)^4 dx = (1/3)∫ u^4 du = (1/3) · u^5/5 + C = (3x + 2)^5 / 15 + C
Example: $\int e^{5x+1} \, dx$
Let u = 5x + 1, du = 5 dx, dx = du/5
∫ e^{5x+1} dx = (1/5)∫ e^u du = (1/5)e^u + C = (1/5)e^{5x+1} + C
Key check: Differentiate your answer to verify. $\frac{d}{dx}[(3x+2)^5/15] = \frac{5(3x+2)^4 \cdot 3}{15} = (3x+2)^4$ ✓
Evaluate $\int_0^1 \frac{x}{\sqrt{x^2 + 1}} \, dx$.
The value of the definite integral.
Let u = x^2 + 1, then du = 2x dx, so x dx = du/2
When x = 0, u = 1; when x = 1, u = 2
∫_0^1 x/√(x^2+1) dx = (1/2)∫_1^2 u^{-1/2} du
= (1/2)[2u^{1/2}]_1^2 = [√u]_1^2 = √2 - 1
$\sqrt{2} - 1 \approx 0.414$.
Evaluate $\int \frac{\ln x}{x} \, dx$.
Answer:
Let $u = \ln x$, then $du = \frac{1}{x} \, dx$.
$\int \frac{\ln x}{x} \, dx = \int u \, du = \frac{u^2}{2} + C = \frac{(\ln x)^2}{2} + C$.
$\int f(g(x))g'(x)\,dx = \int f(u)\,du$
Choose $u =$ "inside" function
Compute $du = g'(x)\,dx$
Change limits: $x=a \to u=g(a)$
With $u = x^2$ and $du = 2x \, dx$, the integral $\int 2x \cos(x^2) \, dx$ becomes $\int \cos(u) \, du = \sin(u) + C = \sin(x^2) + C$. This works because the factor $2x$ is exactly the derivative of $x^2$ — the "inside" function of the composition $\cos(x^2)$. The substitution untangles the chain rule structure, reducing a composite-function integral to a basic one. This is why we call substitution the reverse chain rule: it reverses the process that produced the composite function in the first place.
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
8. Find $\int x^2 \sin(x^3) \, dx$ using substitution. Show all working. 3 MARKS
9. Evaluate $\int_0^{\ln 2} e^x \sqrt{1 + e^x} \, dx$. Show all working. 4 MARKS
10. Explain why $\int e^{x^2} \, dx$ cannot be expressed in terms of elementary functions using substitution. Then explain what mathematicians do when they encounter such integrals in practice (e.g., in statistics). 3 MARKS
1. $u = x^3$, $du = 3x^2 dx$. $\int \cos(u) \, du = \sin(u) + C = \sin(x^3) + C$.
2. $u = 2x + 1$, $du = 2 dx$. $(1/2)\int_1^3 u^3 \, du = (1/2)[u^4/4]_1^3 = (1/8)(81 - 1) = 10$.
3. $u = \sqrt{x}$, $du = \frac{1}{2\sqrt{x}} dx$, so $\frac{dx}{\sqrt{x}} = 2 du$. $\int 2e^u \, du = 2e^u + C = 2e^{\sqrt{x}} + C$.
1. No substitution produces a simpler integral because $2x$ (the derivative of $x^2$) does not appear in the integrand.
2. $u = x^3$, because $3x^2$ (or a constant multiple) appears alongside $e^{x^3}$.
3. $\frac{d}{dx}[-\frac{1}{2}\cos(x^2)] = -\frac{1}{2}(-\sin(x^2)) \cdot 2x = x\sin(x^2)$. ✓ (Note: the original integral should be $\int x \sin(x^2) \, dx$.)
Q8 (3 marks): Let $u = x^3$, $du = 3x^2 \, dx$, so $x^2 \, dx = du/3$ [1]. $\int x^2 \sin(x^3) \, dx = \frac{1}{3}\int \sin(u) \, du$ [1] $= -\frac{1}{3}\cos(u) + C = -\frac{1}{3}\cos(x^3) + C$ [1].
Q9 (4 marks): Let $u = 1 + e^x$, $du = e^x \, dx$ [1]. When $x = 0$, $u = 2$; when $x = \ln 2$, $u = 3$ [1]. $\int_0^{\ln 2} e^x \sqrt{1 + e^x} \, dx = \int_2^3 \sqrt{u} \, du$ [1] $= [\frac{2}{3}u^{3/2}]_2^3 = \frac{2}{3}(3^{3/2} - 2^{3/2}) = \frac{2}{3}(3\sqrt{3} - 2\sqrt{2})$ [1].
Q10 (3 marks): For $\int e^{x^2} \, dx$, any substitution $u = x^2$ requires $du = 2x \, dx$, but the integrand lacks the factor $x$ needed for this substitution [1]. No other elementary substitution works either — this integral has been proven to have no closed-form expression in terms of elementary functions [1]. In practice, mathematicians use numerical methods (e.g., Simpson's rule), series expansions, or special functions like the error function $\text{erf}(x)$ which is defined precisely as $\frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2} \, dt$ [1].
Climb platforms by choosing the right substitution and evaluating transformed integrals. Pool: lesson 8.
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