Jφss Sticks Tuition Centre

Tuition given in the topic of A-Maths Tuition Notes & Tips from the desk of Miss Loi at 7:31 pm (Singapore time)

Updated on 3 November 2008

May the The Temple never see this
(click to enlarge)

With less than ten days (at the time of writing) to go, the skies darkened as the wind picked up, tossing helpless NTUC Fairprice plastic bags up and down like a stock market, as old symptoms (and a couple of new ones) began to surface once more.

While the world deals with the small matter of the possible disintegration of its global financial system, students at Miss Loi’s Temple have far more important things to worry about, like that big A-Maths Integration chapter.

Eh Miss Loi, how come these two Integration questions I never see before in my TYS one?

Latest Edition of Shing Lee Textbook
~~that missed its intended target~~

*Narrowly dodging the big Shing Lee A-Maths textbook that came hurling straight at his face He sat upright and was all ears …

Ooi! Dreaming of your crush again?!
You should know by now that two elements from the New A-Maths Syllabus have impishly made their way into the Integration chapter!

GEN-Y INTEGRATON

A. Integration of 1/(`ax` + `b`)

Starting this year, we have a new entry to the Integration scene:

For x > 0,

int{}{}{1/{ax + b}} dx = {1/a} ln (ax + b) + c

—– (A1)

Having signed a formidable pact with Partial Fractions, the integrals will generally appear in the form of
or
i.e. only linear factors/repeated linear factors in the denominator – very unlikely you’re see quadratic factors like (x²+b²) unless it’s a Hence question with a big clue in the first part.

So whenever you encounter something like the above, regardless of whether you’ve been explicitly told by the question to do so, you MUST perform your Partial Fractions 三大招式 (Three Devastating Moves) to express it in partial fractions first before proceeding to integrate using (A1) above.

SAMPLE QUESTION

Express ${8x+13}/{(1+2x)(2+x)^2}$ in partial fractions. Hence evaluate $int{1}{2}{{8x+13}/{(1+2x)(2+x)^2}} dx$ .
Ans:
Expressing in partial fractions (reference here for the steps if you’re unsure):
1. Check: degree of (8x+13) = 1 < degree of (1+2x)(2+x)² = 3
  ⇒ PROPER (YAY!)
2. ${8x+13}/{(1+2x)(2+x)^2} = A/{1+2x} + B/{2+x} + C/(2+x)^2$
  → 1 x linear factor (1+2x) and
  → 1 x repeated linear factor (2+x)²
3. Multiply both sides by (1+2x)(2+x)²,
  8x+13 = A(2+x)² + B(1+2x)(2+x) + C(1+2x)
  Sub x = -2, -½ … blah blah blah … we get
  ${8x+13}/{(1+2x)(2+x)^2} = 4/{1+2x} - 2/{2+x} + 1/(2+x)^2$
  For all rogue Cover-Up Rule users, note that you can only use it to find the corresponding partial fractions for the terms with linear factors and the higher order of repeated linear factors
  e.g. For this question you can use the Cover-Up rule to find and but NOT .
So,
$int{1}{2}{{8x+13}/{(1+2x)(2+x)^2}} dx = int{1}{2}{4/{1+2x} - 2/{2+x} + 1/(2+x)^2} dx$
${} = 4 int{1}{2}{1/{1+2x}} dx - 2 int{1}{2}{1/{2+x}} dx + int{1}{2}{1/(2+x)^2} dx$
And now you can unleash the expression (in (A1) above) upon these integrals!
At this point, many students will ‘ln ln ln’ till they’ve forgotten how to integrate the 3rd term i.e. the 1/(2+x)².
Remember this?
$int{}{}{(ax+b)^n} dx = 1/a {(ax+b)^{n+1}}/{n+1} + c$
${} = 4 delim{[}{{1/2}ln(1+2x)}{]}^2_1 - 2delim{[}{ln(2+x)}{]}^2_1 - delim{[}{1/{2+x}}{]}^2_1$
Now can you try evaluating $int{2}{3}{{9-4x}/{(2x+3)(x-1)^2}} dx$ ? 😉

B. Integrating Trigonometric Functions Using FURTHER TRIGONOMETRIC IDENTITIES

Basic Rules of Integrating Trigonometric Functions

Firstly, you should all know these by now:

– note the minus sign!
$int{}{}{sec^2 (ax + b)} dx = 1/a tan (ax + b) + c$
$int{}{}{tan^2 (ax + b)} dx$
${}= int{}{}{delim{[}{sec^2 (ax + b) - 1}{]}} dx$ (∵ sec²x = 1 + tan²x)

Starting this year, however, strange forms of trigonometric functions may be pleading for your integrating wizardry in the O-Levels e.g. $int{}{}{sin^2 x} dx$ , , etc.

These are brought to you courtesy of the new Trigonometric Identities of the new syllabus, namely the Double Angle Formulae, Addition Formulae and the Factor Formulae.

While these formulae will be provided in your exam, you’re likely to find yourself frequently using the following derived Double Angle expressions to solve these new wave integration problems: