Integration Revision Notes — 9709 Paper 2

01

The standard integrals

Syllabus requirement · §2.5

“

extend the idea of ‘reverse differentiation’ to include the integration of $\mathrm{e}^{ax+b}$ , $\dfrac{1}{ax+b}$ , $\sin(ax+b)$ , $\cos(ax+b)$ and $\sec^2(ax+b)$ . Knowledge of the general method of integration by substitution is not required.

”

Every one of these is a derivative from Differentiation §06 read backwards. If you know the five derivatives, you already know the five integrals — including the $\dfrac{1}{x}$ case that Paper 1 had to leave out.

The five new integrals

\int \mathrm{e}^{ax+b}\,\mathrm{d}x = \tfrac{1}{a}\mathrm{e}^{ax+b} + c

divide by a

\int \frac{1}{ax+b}\,\mathrm{d}x = \tfrac{1}{a}\ln|ax+b| + c

the missing n = −1 case — note the modulus

\int \sin(ax+b)\,\mathrm{d}x = -\tfrac{1}{a}\cos(ax+b) + c

sine integrates to MINUS cosine

\int \cos(ax+b)\,\mathrm{d}x = \tfrac{1}{a}\sin(ax+b) + c

cosine integrates to plus sine

\int \sec^2(ax+b)\,\mathrm{d}x = \tfrac{1}{a}\tan(ax+b) + c

from the derivative of tan

Three things that get dropped

The $\tfrac{1}{a}$ , every time the inside is $ax + b$ rather than plain $x$ . It is the chain rule's factor coming back out.
The minus in $\int\sin = -\cos$ . Differentiating $\cos$ produces the minus, so integrating $\sin$ must reproduce it.
The modulus in $\ln|ax+b|$ , which is what lets the result exist where $ax+b$ is negative.

No substitution in Paper 2

The syllabus states plainly that "knowledge of the general method of integration by substitution is not required". If an integral seems to need it, you are meant to reach for a trigonometric identity instead (§07) or to recognise a standard form. Substitution belongs to Paper 3.

Worked example9709/22 F/M 2023 Q14 marks

Find the exact value of $\int_{0}^{\frac12\pi} 2\tan^2\!\left(\tfrac12 x\right)\mathrm{d}x.$

Show full working

1
$\tan^2$ is not on the standard list, but $\sec^2$ is. Rearranging $\sec^2 A \equiv 1 + \tan^2 A$ gives $\tan^2 A = \sec^2 A - 1$ : $2\tan^2\!\left(\tfrac12 x\right) = 2\sec^2\!\left(\tfrac12 x\right) - 2$
This first line is a mark on its own. The whole method is “turn it into something on the list”.
2
Integrate. The inside is $\tfrac12 x$ , so $a = \tfrac12$ and you divide by it — which means multiplying by 2: $\int 2\sec^2\!\left(\tfrac12 x\right)\mathrm{d}x = 4\tan\!\left(\tfrac12 x\right)$
3
The $-2$ integrates to $-2x$ , so the antiderivative is $\left[4\tan\!\left(\tfrac12 x\right) - 2x\right]_0^{\pi/2}$ .
4
At $x = \tfrac12\pi$ : $\tfrac12 x = \tfrac14\pi$ and $\tan\tfrac14\pi = 1$ , giving $4(1) - \pi$ .
Radians throughout — tan(π/4) = 1 exactly, which is what makes the answer exact.
5
At $x = 0$ : $4\tan 0 - 0 = 0$ . Subtract.

Answer

$4 - \pi$

Dividing by $a = \tfrac12$ multiplies by 2 — so the coefficient went up, from 2 to 4. Whenever the inside is a fraction of $x$ , expect the constant outside to grow rather than shrink.

Practise the standard integralsReal past-paper questions · Integration of e^(ax+b), 1/(ax+b), sin(ax+b), cos(ax+b), sec^2(ax+b)

Checking your access…

Can you do all of these?

Integrate $\mathrm{e}^{2x}$ and $\dfrac{3}{2x+1}$ , keeping the $\tfrac1a$ and the modulus
Say why $\int\sin(ax+b)$ carries a minus sign
Turn $\cos^2 2x$ into something integrable and evaluate $\int_0^{\pi/4}\cos^2 2x\,\mathrm{d}x$
Use the trapezium rule with four intervals, counting five ordinates
Justify from a sketch whether a trapezium estimate is too big or too small

Integration

The standard integrals

Using trigonometric identities to integrate

The trapezium rule

Can you do all of these?

The standard integrals

The rest of this note

Can you do all of these?