Integrate: choose the method: IB Maths AA SL

Spot it in the exam

Exam phrases to spot (5)

“find $\int (\ldots)\,dx$ ”
“integrate …”
“find the exact value of $\int_1^4 \ldots\,dx$ ”
“using the substitution $u = \ldots$ , or otherwise, find …”
“show that $\int_0^{\frac{\pi}{4}} \ldots\,dx = \ldots$ ”

In disguise (2)

"find an expression for $f(x)$ " when only $f'(x)$ is given and no point: the answer is the whole family, $+c$ included
a part (a) with no integral in it ("show that $f(x) = ax + \frac{b}{x}$ ", or a trig identity): it is the rewrite step, and the integral comes in the next part

Variants (2)

an unknown limit: "given that $\int_1^k \ldots\,dx = \ln 6$ , find $k$ ". The integral becomes an equation in $k$ ; reject any value that the conditions (or the integrand) rule out.
"hence" after a derivative: usually (a) differentiates the inside and (b) integrates a function of it times that derivative, e.g. (a) $\frac{d}{dx}\left(x^3 + 3x\right) = 3\left(x^2 + 1\right)$ , (b) $\int \left(x^2 + 1\right)\left(x^3 + 3x\right)^4 dx = \frac{\left(x^3 + 3x\right)^5}{15} + c$ . Less often, (b) integrates $h'(x)$ or one piece of it, using $\int h'(x)\,dx = h(x) + c$ .

Looks like it, but isn't (4)

a point on the graph is given too: Find f from f′ and a point
values of integrals are given instead of a formula: Definite integrals without the function
an area is asked for: Area between a curve and the x-axis or Area between two curves (an integral can be negative; an area can't:
a product such as $\int xe^x\,dx$ or $\int x\sin x\,dx$ to be found by hand with no "hence": integration by parts, which is HL. A definite integral of one on stays here: row 4 of the table, on the GDC.

Worked example

Split the fraction, then powers and ln

Paper 2 (calculator)

Question

Let $f(x) = \dfrac{6x^2 - \sqrt{x} + 4}{2x}$ , for $x > 0$ .
(a) Show that $f(x) = 3x - \frac{1}{2}x^{-\frac{1}{2}} + 2x^{-1}$ . [2]
(b) Find $\int f(x)\,dx$ . [4]
(c) Find $\int_1^4 f(x)\,dx$ . [2]

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Which method?

Ask these in order, and take the first that fits. The one exception is a bracket to a positive whole-number power: multiplying it out (row 1) always works, but when row 2 or row 3 fits, that row is quicker.

1
Ask
Does it become a sum of standard terms after rewriting: split a fraction over a single term, divide first when a linear top is over a linear bottom (The form p + q/(x − h)), multiply out, write roots and reciprocals as powers, or use an SL identity?
If yes
rewrite, then integrate term by term; $kx^{-1}$ gives $k\ln \lvert x \rvert$
Examples
$\left(x^2 + 1\right)^3 = x^6 + 3x^4 + 3x^2 + 1$ ; $2\sin x\cos x = \sin 2x$ ; $\frac{2x + 7}{x + 3} = 2 + \frac{1}{x + 3}$ , so $2x + \ln \lvert x + 3 \rvert + c$ (row 2 for the second term)
2
Ask
Is every inside linear, $ax + b$ ?
If yes
integrate the outside, then divide by $a$
Examples
$\int (3x + 2)^4\,dx = \frac{(3x + 2)^5}{15} + c$ ; $\int \frac{1}{2x + 5}\,dx = \frac{1}{2}\ln \lvert 2x + 5 \rvert + c$
3
Ask
Is the inside's derivative a factor, up to a constant? A fraction whose top is a multiple of the bottom's derivative counts.
If yes
the reverse chain rule: inspection, or $u = g(x)$
Examples
$\int x\left(x^2 + 1\right)^3 dx = \frac{\left(x^2 + 1\right)^4}{8} + c$ ; $\int \frac{\sin x}{\cos x}\,dx = -\ln \lvert \cos x \rvert + c$
4
Ask
None of these ( $xe^x$ , $x\sin x$ , $e^{x^2}$ , $\frac{1}{x^2 + 1}$ )
If yes
not by hand at SL: only with a "hence" part, or on with the GDC (definite integral)
Examples
$\int_0^1 e^{x^2}\,dx = 1.46$

	Ask	If yes	Examples
1	Does it become a sum of standard terms after rewriting: split a fraction over a single term, divide first when a linear top is over a linear bottom (The form p + q/(x − h)), multiply out, write roots and reciprocals as powers, or use an SL identity?	rewrite, then integrate term by term; $kx^{-1}$ gives $k\ln \lvert x \rvert$	$\left(x^2 + 1\right)^3 = x^6 + 3x^4 + 3x^2 + 1$ ; $2\sin x\cos x = \sin 2x$ ; $\frac{2x + 7}{x + 3} = 2 + \frac{1}{x + 3}$ , so $2x + \ln \lvert x + 3 \rvert + c$ (row 2 for the second term)
2	Is every inside linear, $ax + b$ ?	integrate the outside, then divide by $a$	$\int (3x + 2)^4\,dx = \frac{(3x + 2)^5}{15} + c$ ; $\int \frac{1}{2x + 5}\,dx = \frac{1}{2}\ln \lvert 2x + 5 \rvert + c$
3	Is the inside's derivative a factor, up to a constant? A fraction whose top is a multiple of the bottom's derivative counts.	the reverse chain rule: inspection, or $u = g(x)$	$\int x\left(x^2 + 1\right)^3 dx = \frac{\left(x^2 + 1\right)^4}{8} + c$ ; $\int \frac{\sin x}{\cos x}\,dx = -\ln \lvert \cos x \rvert + c$
4	None of these ( $xe^x$ , $x\sin x$ , $e^{x^2}$ , $\frac{1}{x^2 + 1}$ )	not by hand at SL: only with a "hence" part, or on with the GDC (definite integral)	$\int_0^1 e^{x^2}\,dx = 1.46$

Three that look alike: $\left(x^2 + 1\right)^3$ (row 1: no linear inside and no derivative beside it, so multiply out), $x\left(x^2 + 1\right)^3$ (row 3) and $(2x + 1)^3$ (row 2). The last two both come to a fourth power over $8$ , for different reasons: $\frac{\left(x^2 + 1\right)^4}{8} + c$ and $\frac{(2x + 1)^4}{8} + c$ .

Steps

Ask the table's questions in order,
and stop at the first that fits. Rewrite first where row 1 says so: split a fraction over a single term, divide a linear top by a linear bottom, multiply out, use an identity, write roots and reciprocals as powers.
Integrate term by term,
dividing by $a$ for each linear inside and fixing the constant for each reverse chain.
Indefinite:
add $+c$ . Definite: substitute the upper limit, then the lower, and subtract; on simplify exactly ( $\ln 1 = 0$ , $e^{\ln 2} = 2$ , $\ln 9 = 2\ln 3$ ). On a definite integral comes from the GDC, written in full first (definite integral).
Unknown limit:
set the definite integral, an expression in $k$ , equal to the given value. Solve, keep only the values the conditions allow, and say why the others go.
Check
by differentiating your antiderivative.

Common mistakes

No $+c$
on an indefinite integral: often a lost on its own.
Multiplying by $a$ instead of dividing,
for a linear inside: $\int e^{3x}\,dx = \frac{1}{3}e^{3x} + c$ , not $3e^{3x} + c$ . Differentiating the answer catches it.
Dividing by the inner derivative when the inside isn't linear
$\int e^{x^2}\,dx$ is not $\frac{e^{x^2}}{2x}$ .

More mistakes (7)

The sign of $\int \sin$
$\int \sin x\,dx = -\cos x + c$ .
A right value from a wrong antiderivative
still loses marks: $\int_0^{2\pi} (2 + \sin 3x)\,dx = 4\pi$ , and the wrong antiderivative $2x - 3\cos 3x$ also gives $4\pi$ . Markschemes mark the integrated expression, so its and the final are lost. Check by differentiating, not by the value.
The power rule on $\frac{1}{x}$ ,
which would divide by $0$ : it is $\ln \lvert x \rvert$ . And $\int \frac{1}{x^2}\,dx = -\frac{1}{x} + c$ , not $\ln x^2$ .
Integrating a product factor by factor
multiply out, use the reverse chain rule, or look for a "hence".
A substitution named but not carried out,
or an answer left in $u$ : the answer must be in $x$ .
Keeping a rejected value
of an unknown limit ( $k = -4$ in Example 1.4): .
Degree mode
on with any trig integral, in every type of this guide: check the settings.

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