HL · Further integration 1 · Integration by substitution

Module 38 of 46Lesson 1 of 6 · HL · Further integration 1 · Integration by substitution

Integration by substitution & parts, areas & volumes of revolution · Essential understanding 1

Integration by substitution

IB AA AHL 5.15with Prof. Kitso

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

The big idea: Some integrals look ugly until you rename a piece of them. In substitution you let

u

stand for an inner expression, rewrite the whole integral in terms of

u

, integrate the simpler version, and (for an indefinite integral) put

x

back. For a definite integral you change the limits too, so you never need to go back to

x

By the end of this lesson you can:

choose a good substitution $u = g(x)$ and write $du$ in terms of $dx$
integrate expressions such as $x(x^2+3)^5$ and $x\sqrt{x+4}$
change the limits of a definite integral and finish without going back to $x$
use trigonometric substitutions, for example $u = \cos x$ , $u = \sin x$ or $x = 2\sin\theta$
know when substitution is the right tool, and check an answer on the GDC

🧑🏾 Imagine this…

Kabo is converting a long recipe from cups to millilitres. Instead of doing messy arithmetic with cups, she first replaces one cup by 250 ml everywhere, and then the arithmetic is easy. Substitution does the same job: you replace a messy piece, like

x^2 + 3

, by a single letter

u

. The integral is then written in a new “language” in which it is easy. The only rule is that everything must be translated: the function, the

dx

, and (for definite integrals) the limits.

Substitution

replacing an expression

g(x)

by a new variable

u

so that the integral becomes a standard one.

du

the small change in

u

. From

u = g(x)

we get

\dfrac{du}{dx} = g'(x)

, so

du = g'(x)\,dx

Back-substitution

writing the answer in terms of

x

again by replacing

u

with

g(x)

. Needed for indefinite integrals only.

Changing the limits

for

\int_a^b

, the new limits are

u(a)

and

u(b)

, found by putting

x = a

and

x = b

into

u = g(x)

★ The four steps

1. Choose $u$ . Look for an inner function whose derivative (or a multiple of it) also appears in the integral.
2. Differentiate: $\dfrac{du}{dx} = g'(x)$ , so $du = g'(x)\,dx$ . Solve for $x\,dx$ or $dx$ , whichever you need.
3. Rewrite the whole integral in $u$ . No $x$ may be left (if one is, write it using $u$ : e.g. $u = x + 4 \Rightarrow x = u - 4$ ). For a definite integral change the limits now.
4. Integrate in $u$ . Then either put $u = g(x)$ back (indefinite), or just evaluate between the new limits (definite).

★ Substitutions that show up again and again

$\displaystyle\int x\,f(x^2)\,dx$ : use $u = x^2$ , so $x\,dx = \dfrac12\,du$ .
$\displaystyle\int \frac{f(\ln x)}{x}\,dx$ : use $u = \ln x$ , so $\dfrac{dx}{x} = du$ .
$\displaystyle\int \sin^n x\cos x\,dx$ and $\displaystyle\int \cos^n x\sin x\,dx$ : use $u = \sin x$ and $u = \cos x$ (odd powers: split off one factor and use $\sin^2 x + \cos^2 x = 1$ ).
$\displaystyle\int \tan x\,dx = \int\frac{\sin x}{\cos x}\,dx$ : use $u = \cos x$ , giving $-\ln|\cos x| + C$ .
$\displaystyle\int x\sqrt{x+a}\,dx$ : use $u = x + a$ and replace the extra $x$ by $u - a$ .

⚠️ Watch out

Do not leave $dx$ next to $u$ . $\int 2x(x^2+3)^5\,dx$ becomes $\int u^5\,du$ , not $\int u^5\,dx$ .
In a definite integral, either change the limits to $u$ -values or go back to $x$ before using the old limits. Mixing them ( $u$ in the integrand, $x$ -limits) is the most common lost mark.
A minus sign in $du$ is easy to drop: $u = \cos x \Rightarrow du = -\sin x\,dx$ . Swapping the limits back (to make them increase) also flips the sign.
If the substitution leaves an $x$ you cannot write in terms of $u$ , it was probably the wrong choice. Try a different $u$ , or another method (parts, Lesson 2).

💡 Command terms

“Use the substitution

u = \dots

” means you must show the

du

step and the new limits (M1 for the substitution, A1 for the integrand in

u

, A1 for the limits). “Hence” often points back to a substitution you have just used.

🧪 Try it yourselfArea under y = $\dfrac{x}{\sqrt{}}$ (x + c)

With c = 1 the shaded area from 0 to 3 is exactly

\dfrac{8}{3}

(this is Challenge 1.7). Slide c and watch the area change: the substitution

u = x + c

works for every c, which is why this family of integrals is so common.

Worked examples

Example 1.1The basic move: u = x² + 3

Find

\displaystyle\int x\,(x^2+3)^5\,dx

🧠 Think it through

The inner function is $x^2 + 3$ and its derivative $2x$ is almost there: the integrand has $x$ .
Let $u = x^2 + 3$ . Then $du = 2x\,dx$ , so $x\,dx = \dfrac12\,du$ .
Rewrite, integrate, and put $x$ back.

✍️ By hand
Let , so  and .  (M1)
  (A1)

  (A1)

✅ Answer

\dfrac{(x^2+3)^6}{12} + C

Check by differentiating:

\dfrac{d}{dx}\dfrac{(x^2+3)^6}{12}

= \dfrac{6(x^2+3)^5\cdot 2x}{12}

= x(x^2+3)^5

. ✓

Example 1.2A leftover x: u = x + 4

Find

\displaystyle\int x\sqrt{x+4}\,dx

🧠 Think it through

Here the inner function is $x + 4$ , but its derivative is just 1, so the extra $x$ is not a derivative.
Let $u = x + 4$ . Then $du = dx$ and $x = u - 4$ .
Replace every $x$ by $u - 4$ , multiply out, and integrate term by term.

✍️ By hand
Let , so  and .  (M1)
  (A1)

  (A1)


  (A1)

✅ Answer

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

\quad - \dfrac{8}{3}(x+4)^{\frac{3}{2}} + C

\dfrac{2}{15}(x+4)^{\frac{3}{2}}(3x-8) + C

Both forms are correct. Leave it in the first form in an exam unless the question asks you to factorise.

Example 1.3Definite integrals: change the limits

Use a substitution to find the exact value of
(a)

\displaystyle\int_0^3 \frac{x}{\sqrt{x^2+16}}\,dx

(b)

\displaystyle\int_1^e \frac{(\ln x)^2}{x}\,dx

🧠 Think it through

(a) Let $u = x^2 + 16$ , so $x\,dx = \dfrac12\,du$ . When $x = 0$ , $u = 16$ ; when $x = 3$ , $u = 25$ .
(b) Let $u = \ln x$ , so $du = \dfrac{1}{x}\,dx$ . When $x = 1$ , $u = 0$ ; when $x = e$ , $u = 1$ .
Once the limits are changed, you never go back to $x$ .

✍️ By hand
(a) , . Limits: , .  (M1)(A1)
  (A1)
  (A1)
(b) . Limits: , .  (M1)

  (A1)

🔢 On your TI-Nspire CX II

Check both with the integral template: menu💻 click menu on keypad ▸ 4: Calculus ▸ 2: Numerical Integral. Type the expression, the variable x and the limits.
Menu ▸ Calculus ▸ Numerical Integral gives the integral template

✅ Answer (a)

1

(b)

\dfrac{1}{3}

Example 1.4Trigonometric integrals: tan x and cos³ x

Find the exact value of
(a)

\displaystyle\int_0^{\frac{\pi}{3}}\tan x\,dx

(b)

\displaystyle\int_0^{\frac{\pi}{2}}\cos^3 x\,dx

🧠 Think it through

(a) Write $\tan x = \dfrac{\sin x}{\cos x}$ . The derivative of $\cos x$ is $-\sin x$ , so $u = \cos x$ works.
(b) $\cos^3 x$ has an odd power. Split off one factor: $\cos^3 x = \cos x\,(1 - \sin^2 x)$ , then let $u = \sin x$ .
Careful with the limits and the minus sign in part (a).

✍️ By hand
(a) , . Limits: , .  (M1)(A1)

  (A1)
(b) . Let , . Limits: , .  (M1)(A1)

  (A1)

✅ Answer (a)

\ln 2

(b)

\dfrac{2}{3}

In (a) the answer is positive, as it must be:

\tan x > 0

(0, \dfrac{\pi}{3})

. A negative answer would signal a sign slip.

Example 1.5A trigonometric substitution

Use the substitution

x = 2\sin\theta

to show that

\displaystyle\int_0^2\sqrt{4-x^2}\,dx = \pi

. Explain why this answer makes sense geometrically.

🧠 Think it through

Now the substitution is for $x$ , not for an inner function. $x = 2\sin\theta$ gives $dx = 2\cos\theta\,d\theta$ .
$4 - x^2 = 4 - 4\sin^2\theta = 4\cos^2\theta$ , so $\sqrt{4-x^2} = 2\cos\theta$ (positive, since $\cos\theta \ge 0$ for $0 \le \theta \le \dfrac{\pi}{2}$ ).
Limits: $x = 0 \to \theta = 0$ ; $x = 2 \to \sin\theta = 1 \to \theta = \dfrac{\pi}{2}$ .
You will need $\cos^2\theta = \dfrac{1 + \cos 2\theta}{2}$ .

✍️ By hand
. Limits  and .  (M1)(A1)
  (A1)
  (M1)

  (A1)(AG)
Geometry:  is the upper half of the circle . From  to  the area is a quarter of a circle of radius 2: . ✓

🔢 On your TI-Nspire CX II

Check with the integral template.

✅ Answer

\pi

. It equals the area of a quarter circle of radius 2.

Trigonometric substitutions of this kind are always given in IB questions (“use the substitution x = 2 sin θ”). You do not need to invent them.

Challenge questions

⭐ Challenge 1.6A harder limit changePaper 1 style

Use the substitution

u = 1 - x^2

to show that

\displaystyle\int_0^1 x^3\sqrt{1-x^2}\,dx = \frac{2}{15}

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

$du = -2x\,dx$ , so $x\,dx = -\dfrac12\,du$ .
The integrand has $x^3 = x^2\cdot x$ . Write $x^2 = 1 - u$ , and use $x\,dx$ for the other factor.
Limits: $x = 0 \to u = 1$ , $x = 1 \to u = 0$ . Turning the limits round cancels the minus sign.

✍️ By hand
, , . Limits , .  (M1)(A1)
  (A1)
  (M1)


  (A1)(AG)

✅ Answer

\dfrac{2}{15}

(shown).

⭐ Challenge 1.7Area under y =

\dfrac{x}{\sqrt{}}

(x + 1)Paper 2 style

Let

f(x) = \dfrac{x}{\sqrt{x+1}}

for

x \ge 0

(a) Use the substitution $u = x + 1$ to find the exact area under the graph of $f$ between $x = 0$ and $x = 3$ .
(b) Check your answer using your GDC.
(c) The area between $x = 0$ and $x = k$ is equal to 1. Find $k$ .

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

(a) $u = x + 1$ , so $x = u - 1$ , $dx = du$ . Limits: $0 \to 1$ , $3 \to 4$ .
The integrand becomes $\dfrac{u-1}{\sqrt u} = u^{\frac{1}{2}} - u^{-\frac{1}{2}}$ .
(c) Set the integral from 0 to k equal to 1 and solve with nSolve (a calculator is allowed).

✍️ By hand
(a) , , . Limits  and .  (M1)(A1)

  (A1)

  (A1)

🔢 On your TI-Nspire CX II

(b) Integral template: expression x/√(x+1), variable x, limits 0 and 3.
📍 Where to find it on the TI-Nspire
⌨️ square root: ctrl then x² on the keypad, or type sqrt(
(c) Type nSolve( then the integral template with limits 0 and k, then =1,k=2) and press enter ↵💻 Enter.
📍 Where to find it on the TI-Nspire
nSolve( is in menu💻 click menu on keypad▸3: Algebra▸1: Numerical Solve

✅ Answer (a)

\dfrac83

(b)

2.67

(matches) (c)

k = 1.70

The decimal 2.66667 matches

\dfrac{8}{3}

= 2.6667. On Paper 1 the exact fraction is required; on Paper 2 give 3 s.f.

Next: Lesson 2 · Integration by parts

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL HL · Further integration 1 · Integration by substitution

IB DP Mathematics: Analysis and Approaches (AA) — SL and HL

Integration by substitution

Worked examples

Challenge questions