Integration by substitution

Warm-up

Three quick questions from earlier lessons.

Discover

By the chain rule, $\big(\sin(x^2)\big)' = \cos(x^2) \cdot 2x$ . So what is an antiderivative of $2x\cos(x^2)$ ?

$\sin(x^2)$
$2x\sin(x^2)$

Key idea

Substitution is the chain rule in reverse.

If the integrand has the form

f(g(x))\,g'(x)

, put

u = g(x)

,

du = g'(x)\,dx

:

\displaystyle\int f(g(x))\,g'(x)\,dx = \int f(u)\,du

. If only a constant multiple of

g'(x)

is present, adjust with that constant.

Worked example

$\displaystyle\int x^2 (x^3 + 1)^4\,dx$ .

u = x^3 + 1

,

x^2\,dx = \tfrac{1}{3}du

Inside u = x³ + 1; du = 3x² dx, so x² dx = ⅓ du.

\displaystyle\int \tfrac{1}{3}u^4\,du = \tfrac{1}{15}u^5 + C

Substitute.

\tfrac{1}{15}(x^3 + 1)^5 + C

Back to x.

Your turn, guided

Your turn: $\displaystyle\int 3x^2 e^{x^3}\,dx$ with $u = x^3$ .

e^{x^3} + C

du = 3x² dx exactly.

Practice ★

Find $F$ with $F'(x) = 2x\cos(x^2) + 4$ and $F(0) = 0$ .
Find $F$ with $F'(x) = 2x\cos(x^2) + 8$ and $F(0) = 0$ .
Find $F$ with $F'(x) = 2x\cos(x^2) + 6$ and $F(0) = 0$ .

Watch out!

Practice ★★

Find $F$ with $F'(x) = x\,e^{x^2}$ and $F(0) = 3$ .
Find $F$ with $F'(x) = x\,e^{x^2}$ and $F(0) = 6$ .

Find $F$ with $F'(x) = (5x + 2)^{ 2 }$ and $F(0) = 0$ .
Find $F$ with $F'(x) = (3x + 3)^{ 2 }$ and $F(0) = 0$ .

Exit ticket

Three questions to check you have got it.

Find $F$ with $F'(x) = 2x\cos(x^2) + 4$ and $F(0) = 0$ .
Find $F$ with $F'(x) = 2x\cos(x^2) + 8$ and $F(0) = 0$ .
Find $F$ with $F'(x) = 2x\cos(x^2) + 6$ and $F(0) = 0$ .

Summary card

Substitution

Choose the inside $u = g(x)$ .
Replace $g'(x)\,dx$ by $du$ (adjust a constant).
Integrate in $u$ , then return to $x$ .

Watch out: keep track of constant factors from $du$ .

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