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1. 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 $du$ .

2. Substitution in definite integrals

Warm-up

Three quick questions from earlier lessons.

Discover

In $\displaystyle\int_0^2 2x(x^2 + 1)^3\,dx$ you put $u = x^2 + 1$ . When $x$ goes from 0 to 2, where does $u$ go?

From 1 to 5
From 0 to 2

Key idea

Change the limits with the variable.

\displaystyle\int_a^b f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du

. Then there is no need to go back to

x

Worked example

$\displaystyle\int_0^2 2x(x^2 + 1)^3\,dx$ .

\displaystyle\int_1^5 u^3\,du

u = x² + 1 from 1 to 5.

\dfrac{625 - 1}{4}

u⁴/4 from 1 to 5.

= 156

Compute.

Your turn, guided

Your turn: $\displaystyle\int_0^{\pi} \cos x\,\sin^2 x\,dx$ with $u = \sin x$ .

\displaystyle\int_0^0 u^2\,du = 0

u runs from sin 0 = 0 to sin π = 0.

Practice ★

Evaluate $\displaystyle\int_0^1 2x(x^2 + 1)^{ 4 }\,dx$ .
Evaluate $\displaystyle\int_0^1 2x(x^2 + 1)^{ 3 }\,dx$ .
Evaluate $\displaystyle\int_0^1 2x(x^2 + 1)^{ 2 }\,dx$ .

Watch out!

Practice ★★

Evaluate $\displaystyle\int_0^{\pi/2} \sin^{ 6 } x \cos x\,dx$ .
Evaluate $\displaystyle\int_0^{\pi/2} \sin^{ 3 } x \cos x\,dx$ .

Evaluate $\displaystyle\int_1^{e} \dfrac{(\ln x)^{ 1 } }{x}\,dx$ .

Boss problem. Evaluate $\displaystyle\int_0^{ 8 } x\sqrt{x^2 + 36}\,dx$ with $u = x^2 + 36$ .

Exit ticket

Three questions to check you have got it.

Evaluate $\displaystyle\int_0^1 2x(x^2 + 1)^{ 4 }\,dx$ .
Evaluate $\displaystyle\int_0^1 2x(x^2 + 1)^{ 3 }\,dx$ .
Evaluate $\displaystyle\int_0^1 2x(x^2 + 1)^{ 2 }\,dx$ .

Summary card

Definite integrals by substitution

New limits: $u(a)$ and $u(b)$ .
Integrate in $u$ between them.

Watch out: never mix $x$ -limits with a $u$ -integral.

3. Integrals giving logarithms; exponentials and tangent

Warm-up

Three quick questions from earlier lessons.

Discover

In $\displaystyle\int \dfrac{2x}{x^2 + 1}\,dx$ , the top is the derivative of the bottom. With $u = x^2 + 1$ it becomes $\displaystyle\int \dfrac{du}{u}$ . What is that?

$\ln|u| + C$
$\dfrac{u^0}{0} + C$

Key idea

Three patterns come up again and again.

\displaystyle\int \dfrac{g'(x)}{g(x)}\,dx = \ln|g(x)| + C

;

\displaystyle\int e^{kx}\,dx = \dfrac{e^{kx}}{k} + C

;

\displaystyle\int \tan x\,dx = -\ln|\cos x| + C

Worked example

$\displaystyle\int \dfrac{\cos x}{\sin x + 3}\,dx$ .

u = \sin x + 3

du = \cos x\,dx

The top is the derivative of the bottom.

\ln(\sin x + 3) + C

ln|u|; sin x + 3 > 0.

Your turn, guided

Your turn: $\displaystyle\int e^{5x}\,dx$ .

\dfrac{e^{5x}}{5} + C

Divide by 5.

Practice ★

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

Evaluate $\displaystyle\int_0^1 \dfrac{2x}{x^2 + 2}\,dx$ .
Evaluate $\displaystyle\int_0^1 \dfrac{2x}{x^2 + 9}\,dx$ .

Watch out!

Practice ★★

Find $F$ with $F'(x) = \tan x$ on $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$ and $F(0) = 3$ .
Find $F$ with $F'(x) = \tan x$ on $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$ and $F(0) = -4$ .

Exit ticket

Three questions to check you have got it.

Find $F$ with $F'(x) = e^{ 2x}$ and $F(0) = 0$ .
Find $F$ with $F'(x) = e^{ 5x}$ and $F(0) = 0$ .
Find $F$ with $F'(x) = e^{ 4x}$ and $F(0) = 0$ .

Summary card

Useful patterns

$\dfrac{g'}{g} \to \ln|g|$ .
$e^{kx} \to \dfrac{e^{kx}}{k}$ .
$\tan x \to -\ln|\cos x|$ .

Watch out: a minus sign from $du$ .

4. Algebra before integrating: division and completing the square

Warm-up

Three quick questions from earlier lessons.

Discover

$\dfrac{1}{x^2 + 4x + 5}$ has no obvious inside function. Which rewriting helps?

$\dfrac{1}{(x + 2)^2 + 1}$
$\dfrac{1}{x^2} + \dfrac{1}{4x} + \dfrac{1}{5}$

Key idea

Rewrite first, then integrate.

Divide when the top has a degree at least that of the bottom, or the bottom is one term:

\dfrac{x^2 + 1}{x} = x + \dfrac{1}{x}

. Complete the square to reach

\displaystyle\int \dfrac{du}{u^2 + a^2} = \dfrac{1}{a}\arctan\dfrac{u}{a} + C

Worked example

$\displaystyle\int \dfrac{dx}{x^2 + 6x + 13}$ .

x^2 + 6x + 13 = (x + 3)^2 + 4

x² + 6x + 9 + 4.

\displaystyle\int \dfrac{du}{u^2 + 4}

u = x + 3, a = 2.

\tfrac{1}{2}\arctan\dfrac{x + 3}{2} + C

The arctan formula.

Your turn, guided

Your turn: $\displaystyle\int \dfrac{x^2 + 2}{x}\,dx$ for $x > 0$ .

\dfrac{x^2}{2} + 2\ln x + C

x + 2/x.

Practice ★

Find $F$ with $F'(x) = \dfrac{1}{x^2 + 9}$ and $F(0) = 0$ .
Find $F$ with $F'(x) = \dfrac{1}{x^2 + 4}$ and $F(0) = 0$ .

Evaluate $\displaystyle\int_1^2 \dfrac{x^2 + 5x + 1}{x}\,dx$ .
Evaluate $\displaystyle\int_1^2 \dfrac{x^2 + 4x + 1}{x}\,dx$ .

Watch out!

Practice ★★

Find $F$ with $F'(x) = \dfrac{1}{x^2 + 6x + 10}$ and $F(-3) = 0$ .
Find $F$ with $F'(x) = \dfrac{1}{x^2 + 8x + 17}$ and $F(-4) = 0$ .

Exit ticket

Three questions to check you have got it.

Find $F$ with $F'(x) = \dfrac{1}{x^2 + 9}$ and $F(0) = 0$ .
Find $F$ with $F'(x) = \dfrac{1}{x^2 + 4}$ and $F(0) = 0$ .
Find $F$ with $F'(x) = \dfrac{1}{x^2 + 36}$ and $F(0) = 0$ .

Summary card

Algebra first

Divide term by term when the bottom is one term.
Complete the square for $\dfrac{1}{x^2 + bx + c}$ , then arctan.

Watch out: $\dfrac{1}{a + b} \ne \dfrac{1}{a} + \dfrac{1}{b}$ .

Answers: Techniques: substitution

1. Integration by substitution

$4 x + \sin{\left(x^{2} \right)}$
$8 x + \sin{\left(x^{2} \right)}$
$6 x + \sin{\left(x^{2} \right)}$

$\frac{e^{x^{2}}}{2} + \frac{5}{2}$
$\frac{e^{x^{2}}}{2} + \frac{11}{2}$

$\frac{\left(5 x + 2\right)^{3}}{15} - \frac{8}{15}$
$\frac{\left(3 x + 3\right)^{3}}{9} - 3$

$4 x + \sin{\left(x^{2} \right)}$
$8 x + \sin{\left(x^{2} \right)}$
$6 x + \sin{\left(x^{2} \right)}$

2. Substitution in definite integrals

$\frac{31}{5}$
$\frac{15}{4}$
$\frac{7}{3}$

$\frac{1}{7}$
$\frac{1}{4}$

$\frac{1}{2}$

$\frac{784}{3}$

$\frac{31}{5}$
$\frac{15}{4}$
$\frac{7}{3}$

3. Integrals giving logarithms; exponentials and tangent

$\frac{e^{2 x}}{2} - \frac{1}{2}$
$\frac{e^{5 x}}{5} - \frac{1}{5}$

$\ln{\left(\frac{3}{2} \right)}$
$\ln{\left(\frac{10}{9} \right)}$

$3 - \ln{\left(\cos{\left(x \right)} \right)}$
$- \ln{\left(\cos{\left(x \right)} \right)} - 4$

$\frac{e^{2 x}}{2} - \frac{1}{2}$
$\frac{e^{5 x}}{5} - \frac{1}{5}$
$\frac{e^{4 x}}{4} - \frac{1}{4}$

4. Algebra before integrating: division and completing the square

$\frac{\tan^{-1}{\left(\frac{x}{3} \right)}}{3}$
$\frac{\tan^{-1}{\left(\frac{x}{2} \right)}}{2}$

$\ln{\left(2 \right)} + \frac{13}{2}$
$\ln{\left(2 \right)} + \frac{11}{2}$

$\tan^{-1}{\left(x + 3 \right)}$
$\tan^{-1}{\left(x + 4 \right)}$

$\frac{\tan^{-1}{\left(\frac{x}{3} \right)}}{3}$
$\frac{\tan^{-1}{\left(\frac{x}{2} \right)}}{2}$
$\frac{\tan^{-1}{\left(\frac{x}{6} \right)}}{6}$