Edexcel IAL - Pure Maths 3- 5.2 Integration by recognition of known derivatives- Study notes - New syllabus
Edexcel IAL – Pure Maths 3- 5.2 Integration by recognition of known derivatives -Study notes- New syllabus
Edexcel IAL – Pure Maths 3- 5.2 Integration by recognition of known derivatives -Study notes -Edexcel A level Physics – per latest Syllabus.
Key Concepts:
- 5.2 Integration by recognition of known derivatives
Integration by Recognition of Known Derivatives
Some integrals can be evaluated directly by recognising them as the reverse of differentiation.
This method avoids substitution by identifying a function and its derivative within the integrand.
Standard Recognition Results
Logarithmic Form
If the integrand is of the form:
\( \dfrac{f'(x)}{f(x)} \)
then:
\( \displaystyle \int \dfrac{f'(x)}{f(x)}\,dx = \ln|f(x)| + C \)
Power Form
If the integrand is of the form:
\( f'(x)[f(x)]^n \)
where \( n \ne -1 \), then:
\( \displaystyle \int f'(x)[f(x)]^n\,dx = \dfrac{[f(x)]^{n+1}}{n+1} + C \)
Key Idea
Look for a function and its derivative appearing together.
Common Trigonometric Integrals by Recognition
- \( \tan x = \dfrac{\sin x}{\cos x} \Rightarrow \ln|\cos x| \)
- \( \sec^2 x = \dfrac{d}{dx}(\tan x) \)
- \( \dfrac{d}{dx}(\sin x) = \cos x \)
- \( \dfrac{d}{dx}(\cos x) = -\sin x \)
Use of Trigonometric Identities
Some integrals require identities before recognition can be applied.
Common identities:
\( \sin^2 x = \dfrac{1 – \cos 2x}{2} \)
\( \cos^2 x = \dfrac{1 + \cos 2x}{2} \)
\( \tan^2 x = \sec^2 x – 1 \)
Example
Find:
\( \displaystyle \int \tan x\,dx \)
▶️ Answer / Explanation
Write:
\( \tan x = \dfrac{\sin x}{\cos x} \)
Since:
\( \dfrac{d}{dx}(\cos x) = -\sin x \)
The integral becomes:
\( -\ln|\cos x| + C \)
Answer:
\( \ln|\sec x| + C \)
Example
Find:
\( \displaystyle \int \sec^2 2x\,dx \)
▶️ Answer / Explanation
Recognise that:
\( \dfrac{d}{dx}(\tan 2x) = 2\sec^2 2x \)
So:
\( \int \sec^2 2x\,dx = \dfrac{1}{2}\tan 2x + C \)
Example
Find:
\( \displaystyle \int \left(\sin^2 x + \tan^2 x + \cos^2 3x\right)\,dx \)
▶️ Answer / Explanation
Use identities:
\( \sin^2 x = \dfrac{1 – \cos 2x}{2} \)
\( \tan^2 x = \sec^2 x – 1 \)
\( \cos^2 3x = \dfrac{1 + \cos 6x}{2} \)
Substitute and integrate term by term:
\( \int \left(\dfrac{1}{2} – \dfrac{1}{2}\cos 2x + \sec^2 x – 1 + \dfrac{1}{2} + \dfrac{1}{2}\cos 6x\right) dx \)
\( = \int \left(\sec^2 x – \dfrac{1}{2}\cos 2x + \dfrac{1}{2}\cos 6x\right) dx \)
\( = \tan x – \dfrac{1}{4}\sin 2x + \dfrac{1}{12}\sin 6x + C \)