HL · Further integration 1 · Integrating more functions

Module 39 of 47Lesson 1 of 8 · HL · Further integration 1 · Integrating more functions

Further integration & kinematics · Essential understanding 1

Integrating more functions

IB AI AHL 5.11with 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: Integration undoes differentiation. Now that you can differentiate powers like

\sqrt{x}

\sin x

\cos x

e^x

and

\ln x

, you can reverse every one of those rules, and also integrate a function of a linear expression such as

\cos(3x-1)

e^{2-5x}

By the end of this lesson you can:

integrate $x^n$ for any rational $n$ except $-1$ , including roots and negative powers
integrate $\sin x$ , $\cos x$ , $\dfrac{1}{\cos^2 x}$ , $\dfrac{1}{x}$ and $e^x$
integrate these when $x$ is replaced by a linear expression $ax+b$
find the constant of integration from a point on the curve
check an answer by differentiating it, or with the integral template on your GDC

🧑🏾 Imagine this…

Think of a lift. Pressing the up button and then the down button puts you back where you started. Differentiating is one button and integrating is the other. If you know that

\sin x

turns into

\cos x

when you differentiate it, then going backwards,

\cos x

must come from

\sin x

. Every integration rule in this lesson is just a differentiation rule read backwards.

Antiderivative

a function

F(x)

whose derivative is the function you were given:

F'(x) = f(x)

Indefinite integral

\displaystyle\int f(x)\,dx = F(x) + C

. The answer is a whole family of curves, so we always add the constant of integration

C

Rational power

a power that is a fraction or negative, such as

x^{\frac{1}{2}} = \sqrt{x}

x^{-3} = \dfrac{1}{x^3}

★ The integrals to know (all with + C)

$\displaystyle\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C$ for $n \ne -1$ (any rational $n$ ) (in the IB booklet for integer $n$ ; use it for fractions too)
$\displaystyle\int \sin x\,dx = -\cos x + C$ $\displaystyle\int \cos x\,dx = \sin x + C$ (x in radians)
$\displaystyle\int \dfrac{1}{\cos^2 x}\,dx = \tan x + C$ $\displaystyle\int e^x\,dx = e^x + C$ $\displaystyle\int \dfrac{1}{x}\,dx = \ln|x| + C$
Linear inside: $\displaystyle\int f(ax+b)\,dx = \dfrac{1}{a}F(ax+b) + C$ . Divide by the number multiplying $x$ .
Constants come out front: $\displaystyle\int k\,f(x)\,dx = k\displaystyle\int f(x)\,dx$ . Sums are integrated term by term.

Rewrite each term as a number times a power: $\sqrt{x} = x^{\frac{1}{2}}$ , $\dfrac{1}{x^3} = x^{-3}$ .
Raise the power by 1, then divide by the new power.
For $f(ax+b)$ also divide by $a$ .
Add + C and, if you can, check by differentiating.

⚠️ Watch out

Do not use the power rule on $\dfrac{1}{x}$ . Raising $x^{-1}$ by one gives $x^0$ and a division by zero. The answer is $\ln|x|$ .
Trig signs. Differentiating $\cos x$ gives $-\sin x$ , so $\displaystyle\int \sin x\,dx = \mathbf{-}\cos x$ . It is the integral of sin that has the minus.
Divide, do not multiply, by the number in front of x. $\displaystyle\int e^{3x}\,dx = \dfrac{1}{3}e^{3x}$ , not $3e^{3x}$ .
Always set your calculator to radians for anything with sin, cos or tan inside an integral.

💡 Command terms

Find or integrate: give the antiderivative with + C. Show that: write every step so the printed answer appears. Hence: use the previous part. In a definite integral there is no + C.

🧪 Try it yourselfDoes the gradient of the integral match the integrand?

The blue curve is

F(x) = \dfrac{\sin(ax)}{a}

, our claimed integral of

\cos(ax)

(the red curve). Drag the tangent along the blue curve. Its gradient should equal the height of the red curve at the same x. Try a = 2, then change a. This is exactly why we divide by a.

Worked examples

Example 1.1Roots and negative powers

Find

\displaystyle\int \left(4\sqrt{x} - \dfrac{10}{x^3} + 7\right)dx

🧠 Think it through

Rewrite every term as a power: $4x^{\frac{1}{2}} - 10x^{-3} + 7x^0$ .
Raise each power by one and divide by the new power.
Check by evaluating a definite integral on the GDC and comparing with the antiderivative.

✍️ By hand

\displaystyle\int \left(4x^{\frac{1}{2}} - 10x^{-3} + 7\right)dx

= 4\cdot\dfrac{x^{\frac{3}{2}}}{\frac{3}{2}} - 10\cdot\dfrac{x^{-2}}{-2} + 7x + C

= \dfrac{8}{3}x^{\frac{3}{2}} + 5x^{-2} + 7x + C = \dfrac{8}{3}x\sqrt{x} + \dfrac{5}{x^2} + 7x + C

🔢 On your TI-Nspire CX II

Check between x = 1 and x = 4. First the integral template, then the antiderivative $F(4) - F(1)$ . The two must agree.
Same number both ways

✅ Answer

\dfrac{8}{3}x^{\frac{3}{2}} + \dfrac{5}{x^2} + 7x + C

Dividing by

\dfrac{3}{2}

is the same as multiplying by

\dfrac{2}{3}

. Do it on paper: 4 ×

\dfrac{2}{3}

\dfrac{8}{3}

Example 1.2Trig, exponential and

\dfrac{1}{x}

together

(a) Find

\displaystyle\int \left(5\cos x + 2\sin x + 3e^x - \dfrac{4}{x}\right)dx

.
(b) Find

\displaystyle\int \dfrac{6}{\cos^2 x}\,dx

🧠 Think it through

Take each term separately, with its own rule from the key box.
Watch the sign for $\int \sin x$ . It becomes $-\cos x$ , and $2\times(-\cos x) = -2\cos x$ .
$\dfrac{6}{\cos^2 x} = 6\times\dfrac{1}{\cos^2 x}$ , and $\dfrac{1}{\cos^2 x}$ integrates to $\tan x$ .

✍️ By hand

(a)

\displaystyle\int 5\cos x\,dx = 5\sin x

\displaystyle\int 2\sin x\,dx = -2\cos x

\displaystyle\int 3e^x\,dx = 3e^x

\displaystyle\int \dfrac{4}{x}\,dx = 4\ln|x|

So the answer is

5\sin x - 2\cos x + 3e^x - 4\ln|x| + C

(b)

\displaystyle\int \dfrac{6}{\cos^2 x}\,dx = 6\tan x + C

🔢 On your TI-Nspire CX II

Check (a) between 1 and 2 with the antiderivative. Set the calculator to radians first: doc ▾💻 File menu ▸ Settings ▸ Document Settings ▸ Angle: Radian.
A match, so the antiderivative is right

✅ Answer (a)

5\sin x - 2\cos x + 3e^x - 4\ln|x| + C

(b)

6\tan x + C

Example 1.3Functions of a linear expression

Find (a)

\displaystyle\int \sin(4x)\,dx

(b)

\displaystyle\int e^{3-2x}\,dx

(c)

\displaystyle\int \dfrac{6}{3x+2}\,dx

(d)

\displaystyle\int \dfrac{1}{\cos^2(2x)}\,dx

🧠 Think it through

For $f(ax+b)$ , integrate as if it were just $x$ , then divide by $a$ , the number in front of x.
In (b) the number in front of x is $-2$ , so we divide by $-2$ .
In (c) pull the 6 out, then use the $\ln$ rule and divide by 3.

✍️ By hand

(a)

-\dfrac{1}{4}\cos(4x) + C

(b)

\dfrac{e^{3-2x}}{-2} + C = -\dfrac{1}{2}e^{3-2x} + C

(c)

6\cdot\dfrac{1}{3}\ln|3x+2| + C = 2\ln|3x+2| + C

(d)

\dfrac{1}{2}\tan(2x) + C

🔢 On your TI-Nspire CX II

Check (b) from x = 0 to 1: the exact value is $\dfrac{1}{2}(e^3 - e)$ . Then check (c).
(b) and (c) confirmed

✅ Answer (a)

-\dfrac{1}{4}\cos 4x + C

(b)

-\dfrac{1}{2}e^{3-2x} + C

(c)

2\ln|3x+2| + C

(d)

\dfrac{1}{2}\tan 2x + C

A quick sanity check: differentiate your answer. For (b), the chain rule gives −½ × e^(3−2x) × (−2) = e^(3−2x). It works.

Example 1.4Roots and negative powers of a linear expression

Find (a)

\displaystyle\int \sqrt{3x+1}\,dx

(b)

\displaystyle\int \dfrac{1}{(2x-1)^3}\,dx

(c)

\displaystyle\int \dfrac{8}{5-2x}\,dx

.
(d) Hence evaluate

\displaystyle\int_1^2 \dfrac{1}{(2x-1)^3}\,dx

🧠 Think it through

(a) $\sqrt{3x+1} = (3x+1)^{\frac{1}{2}}$ . Use the power rule, then divide by 3.
(b) Write $(2x-1)^{-3}$ , raise the power to $-2$ , divide by $-2$ , then divide by 2.
(c) This is $\dfrac{1}{x}$ in disguise: $\ln|5-2x|$ , divided by $-2$ .
(d) Put the limits into your answer to (b): top minus bottom.

✍️ By hand

(a)

\dfrac{(3x+1)^{\frac{3}{2}}}{\frac{3}{2}}\cdot\dfrac{1}{3} + C = \dfrac{2}{9}(3x+1)^{\frac{3}{2}} + C

(b)

\dfrac{(2x-1)^{-2}}{-2}\cdot\dfrac{1}{2} + C = -\dfrac{1}{4(2x-1)^2} + C

(c)

8\cdot\dfrac{\ln|5-2x|}{-2} + C = -4\ln|5-2x| + C

(d)

\left[-\dfrac{1}{4(2x-1)^2}\right]_1^2 = -\dfrac{1}{36} - \left(-\dfrac{1}{4}\right) = \dfrac{8}{36} = \dfrac{2}{9}

🔢 On your TI-Nspire CX II

Check (a) from 0 to 5 (exact: $\dfrac{2}{9}(16^{\frac{3}{2}} - 1) = 14$ ) and (d).

✅ Answer (a)

\dfrac{2}{9}(3x+1)^{\frac{3}{2}} + C

(b)

-\dfrac{1}{4(2x-1)^2} + C

(c)

-4\ln|5-2x| + C

(d)

\dfrac{2}{9}

Notice in (c) the minus sign in front of the 2x. Dividing by −2 turns the answer negative. Many students lose a mark here.

Example 1.5A curve through a point (radians!)

The gradient of a curve is

\dfrac{dy}{dx} = 6e^{2x} + 8\cos(2x)

. The curve passes through the point

(0, 5)

(a) Find the equation of the curve.
(b) Find the value of y when $x = \dfrac{\pi}{4}$ .

🧠 Think it through

Integrate each term, dividing by 2 for the linear inside part.
Use $x = 0,\ y = 5$ to find C. Remember $\sin 0 = 0$ and $e^0 = 1$ .
For (b) the angle is in radians, so set the GDC to radians before you type the value.

✍️ By hand

y = \dfrac{6e^{2x}}{2} + \dfrac{8\sin(2x)}{2} + C = 3e^{2x} + 4\sin(2x) + C

(0, 5)

5 = 3(1) + 0 + C

, so

C = 2

y = 3e^{2x} + 4\sin(2x) + 2

x = \dfrac{\pi}{4}

y = 3e^{\frac{\pi}{2}} + 4\sin\dfrac{\pi}{2} + 2 = 3e^{\frac{\pi}{2}} + 6 \approx 20.4

🔢 On your TI-Nspire CX II

Define the function in radian mode and test the point and the value.
y(0) = 5 confirms C = 2

✅ Answer (a)

y = 3e^{2x} + 4\sin 2x + 2

(b)

y \approx 20.4

Challenge questions

⭐ Challenge 1.6Mixed integralPaper 1 style

Find

\displaystyle\int \left(6\sqrt{x} + \dfrac{5}{x^2} + 2\cos(3x)\right)dx

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

Show the full solution

🧠 Think it through

Rewrite as $6x^{\frac{1}{2}} + 5x^{-2} + 2\cos 3x$ .
Integrate term by term. For the cosine, divide by 3.
Finish with + C.

✍️ By hand

6\cdot\dfrac{x^{\frac{3}{2}}}{\frac{3}{2}} + 5\cdot\dfrac{x^{-1}}{-1} + 2\cdot\dfrac{\sin 3x}{3} + C

🔢 On your TI-Nspire CX II

Check between 1 and 2 (radians).

✅ Answer

4x^{\frac{3}{2}} - \dfrac{5}{x} + \dfrac{2}{3}\sin 3x + C

⭐ Challenge 1.7From gradient to curve to areaPaper 2 style

A curve has gradient

\dfrac{dy}{dx} = \dfrac{4}{2x+1} + 6e^{-x}

, for

x > 0

. The curve passes through

(0, 2)

(a) Find the equation of the curve.
(b) Find y when x = 2.
(c) Evaluate $\displaystyle\int_0^2 \left(\dfrac{4}{2x+1} + 6e^{-x}\right)dx$ , giving an exact answer.
(d) Explain how your answers to (b) and (c) are connected.

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

Show the full solution

🧠 Think it through

$\dfrac{4}{2x+1}$ integrates to $4\times\dfrac{1}{2}\ln(2x+1)$ . And $e^{-x}$ integrates to $\dfrac{e^{-x}}{-1}$ .
Use the point (0, 2) to find C. Remember ln 1 = 0 and e⁰ = 1.
A definite integral of the gradient is the change in y.

✍️ By hand
(a) . At : , so .
(b) 
(c) 
(d) : the integral of the gradient is the change in y.

🔢 On your TI-Nspire CX II

Check (b) and (c) together.

✅ Answer (a)

y = 2\ln(2x+1) - 6e^{-x} + 8

(b)

10.4

(c)

2\ln 5 - 6e^{-2} + 6\ (\approx 8.41)

(d) The integral from 0 to 2 equals

y(2) - y(0)

, the change in y.

Next: Lesson 2 · Integration by inspection

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL HL · Further integration 1 · Integrating more functions

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL

Integrating more functions

Worked examples

Challenge questions