Six formulas for quadratic denominators, and completing the square to reach them

What to assume they know

• The standard formulae, including the inverse sine and inverse tangent rows, from module m01
• Substitution, from the first topic of this module
• Completing the square on a quadratic expression, from Class IX or X
• The identity relating the square of the secant to the square of the tangent, and the identity relating the square of the cosine to the square of the sine
• The integral of the secant, derived in the first topic of this module
• Comparing coefficients on two sides of a polynomial identity

What they should be able to do

• Recite the six formulas as three related pairs and say what distinguishes each pair
• Say which of the six are proved by splitting into partial fractions and which by a trigonometric substitution, and name the substitution in each case
• Complete the square on a general quadratic and read off which of the six shapes results
• Handle a leading coefficient other than one, and say where the extra factor goes
• Split a linear numerator into a multiple of the denominator's derivative plus a constant, and solve for the two unknowns
• Reduce any of the four derived types to one of the six formulas
• Decide, from an integrand, which of the six is the target before starting work
• Recognise the forward reference the chapter drops into the middle of its proofs and say where it resolves

Where it usually goes wrong

• "There are six formulas to memorise." There are three, each in two versions, bare and under a root; and of the six answers only two are inverse functions. Sorting them that way turns a memory list into a decision with two branches.
• "Completing the square is optional if the quadratic factorises." It is not optional for formulas (3), (5) and (6), whose denominators never factorise over the reals. It is optional for (1) and (2), and taking the factorising route there is what §7.5 will do instead.
• "The leading coefficient can be ignored." It comes out of the integral as a reciprocal — or as a reciprocal square root when the quadratic is under a root — and Example 9 (ii) and (iii) exist to show the two cases. Students who complete the square without pulling the coefficient out first get a shifted square with a wrong scale.
• "The two unknowns in the linear-numerator split are found by substituting values." The chapter compares coefficients, which is the method that always works; substituting convenient values is a shortcut that works for the factorising cases of §7.5 and is not what type (9) does.
• "A linear numerator means the answer will be a logarithm." It means the answer will usually be a logarithm plus an inverse function, because the split produces two integrals of different kinds. Example 10 has one of each in both parts.
• "Formula (4) and formula (6) are the same formula." They differ in the sign under the root, and therefore in which trigonometric substitution proves them — a secant for one and a tangent for the other — even though their printed answers look almost identical. That near-identity is a trap, not a simplification.
• "§7.4 stands on its own." Four of its six proofs are substitutions and two of those need the integral of the secant from §7.3.1. It is the second storey of the same building.
• "The forward reference means the proof of (1) is incomplete." The proof is complete; only the name of the technique is deferred. Say which is deferred, or students will believe a gap has been left.

Questions to check understanding

• Match a given integrand to one of the six formulas and evaluate it
• Complete the square on a general quadratic and identify which shape results
• Integrate with a leading coefficient other than one, both with and without a root — the form of Exercise 7.4 Q11 and Q18
• Split a linear numerator and solve for the two unknowns by comparing coefficients
• Evaluate a type (9) or type (10) integral end to end — the form of Example 10 and of Exercise 7.4 Q16 to Q23
• Choose the correct integral from four options — the form of Exercise 7.4 Q24 and Q25
• Name the substitution that proves a stated one of the six formulas

Examples worth working on the board

Values marked verified are worked out here from the chapter's own printed data; neither answers file was opened, and this chapter prints no answers to its exercises.

• The six formulas (§7.4, Part II pp. 243–244), numbered (1) to (6). Read off the page images, with every radical and modulus bar checked. They fall into three pairs., because the chapter's own numbering hides the structure: bare denominator, difference and sum of squares — (1) and (2), both answering with half the reciprocal of twice the constant times a logarithm of a ratio, one with the variable's square leading and one with the constant's; bare denominator, sum of squares — (3), answering with the inverse tangent scaled by the reciprocal of the constant; the same three under a square root — (4), (5) and (6), answering with a logarithm, an inverse sine and a logarithm respectively. Note that (3) and (5) are the two that give an inverse function and the other four give logarithms, which is a better sorting principle for a student than the printed order.
• The proofs (§7.4, Part II pp. 244–246). Two of the six are algebraic and four are substitutions. (1) splits the reciprocal of a difference of two squares into two simple fractions by writing the numerator as a difference of the two factors; (2) says it follows the same way. (3) substitutes the variable equal to the constant times a tangent; (4) uses a secant; (5) uses a sine; (6) uses a tangent again. All four substitutions collapse the denominator by a Pythagorean identity, and (4) and (6) then need the integral of the secant, which the first topic of this module derived. Say that out loud: §7.4 is not independent of §7.3.1, it is built on it.
• The Note in the middle of the proofs (Part II p. 245). One boxed sentence saying the technique used in the first proof will be explained in §7.5. Verified: §7.5 is Integration by Partial Fractions and it begins on Part II p. 252, so the reference resolves correctly in this edition. Worth showing, because it is the chapter telling the reader that a technique is being borrowed before it is taught, and because most of this chapter's forward references are not this tidy.
• The four derived types (§7.4, Part II pp. 246–247), numbered (7) to (10). (7) and (8) are a general quadratic downstairs, bare and under a root; (9) and (10) are the same two with a linear numerator. (7) prints the completing-the-square identity in full and then names the sign of the leftover constant as the thing that decides which of the six you land on. (9) prints the splitting identity: the linear numerator is written as an unknown multiple of the denominator's derivative plus a second unknown, and the two are found by comparing coefficients. (8) and (10) are each described as the previous one done under a root. Read off the page images.
• Example 8 (Part II p. 247), two parts. (i) A difference of the variable's square and sixteen, matched to formula (1) after writing sixteen as a square. Verified: the answer is an eighth of the logarithm of the modulus of the ratio of the variable less four to the variable plus four. (ii) Twice the variable less its square, under a root: completing the square turns it into one less a shifted square, and formula (5) applies. Verified: the answer is the inverse sine of the variable less one. Both marginal citations on this page are in arabic numerals and both are correct.
• Example 9 (Part II pp. 247–249), three parts and the heart of the topic. (i) A monic quadratic downstairs: complete the square, substitute, use (3). Verified: the answer is half the inverse tangent of half the shifted variable. (ii) A leading coefficient of three: the chapter pulls the three out first, then completes the square inside, then uses (1) on a difference of squares. Verified: the answer is the reciprocal of seventeen times the logarithm of the modulus of the ratio of three times the variable less two to the variable plus five, after two constants have been folded together. The folding is printed in full and is a good use of module m01's Note about a single constant. See the note below about the marginal citation on this page. (iii) A leading coefficient of five, under a root: the five comes out as a reciprocal square root, the square is completed inside, and (4) applies. Verified: the answer is the reciprocal of the square root of five times a logarithm of a modulus.
• Example 10 (Part II pp. 249–251), two parts, both with a linear numerator. (i) The variable plus two, over twice its square plus six times it plus five. The split gives a quarter and a half as the two unknowns; the first piece integrates by substituting for the whole denominator, the second by completing the square and using (3). Verified by working added here: the two unknowns are a quarter and a half, and the answer is a quarter of the logarithm of the modulus of the quadratic plus half the inverse tangent of twice the variable plus three. (ii) The variable plus three, over the root of five less four times the variable less its square. The two unknowns are minus a half and one. Verified: the answer is minus the root of the quadratic plus the inverse sine of the ratio of the variable plus two to three.
• Exercise 7.4 (Part II pp. 251–252), twenty-five items, of which the last two are multiple choice. Grouped by shape — an added classification, not the chapter's: Q1 to Q9 are direct hits on one of the six after a substitution that turns a higher power into a square; Q10 to Q15 need the square completed with a leading coefficient of one; Q16 to Q23 have a linear numerator and are the type (9) and type (10) split; and Q11, Q18 and Q23 combine a leading coefficient other than one with a linear numerator, which is the hardest shape in the set. Q7 is worth a separate mention: its numerator is linear and its denominator is a root, so it is type (10) with the smallest possible numbers.
• Exercise 7.4, questions 24 and 25 (Part II p. 252). Two multiple-choice items. Verified by working added here: Q24's denominator completes to a shifted square plus one, so formula (3) with unit constant gives the inverse tangent of the shifted variable, which is the second option. Q25's denominator, under the root, completes to a constant less a shifted square scaled by four; carrying the four out gives an inverse sine of a ratio whose denominator is nine and whose numerator is eight times the variable less nine, halved — the second option. The chapter prints no answers.
• The Summary (Part II p. 289 and Part II p. 290). Part II p. 289 reprints the six formulas as (i) to (vi) under a heading naming them special functions; Part II p. 290 carries the completing-the-square identity and the numerator-splitting identity as items (iv) and (v) under a different heading. The six and the four are therefore separated in the Summary by two pages and three bullets, which is worth knowing when planning revision, since the six are useless without the four.

Figures to have open

• A three-pair layout of the six formulas for section 2, and a reduction table of the four derived types for section 7, and a shape grouping of Exercise 7.4 for section 11. All three carry the chapter's own content, from Part II pp. 243–247 and Part II pp. 251–252; the arrangements are added here. Build all three with the repo's DataTable component.
• A decision tree for section 5: complete the square, look at the sign of what is left, branch to one of three shapes. Not in the book; the chapter states the branch in one clause on Part II p. 246 and draws nothing.
• No figure is available from the chapter for any section of this topic. The whole chapter prints one figure, on Part II p. 267, and it belongs to module m03. Every picture listed here is an added construction over the chapter's own content.

Where this sits in the book

• NCERT Class 12 Mathematics, Chapter 7 "Integrals", §7.4 Integrals of Some Particular Functions, the six formulas, Part II pp. 243–244
• The six proofs and the boxed Note forward-referencing §7.5, Part II pp. 244–246
• The four derived types (7) to (10), Part II pp. 246–247
• Examples 8, 9 and 10, Part II pp. 247–251
• Exercise 7.4, questions 1 to 25, Part II pp. 251–252
• Summary, Integrals of some special functions, Part II p. 289, and Some special types of integrals items (iv) and (v), Part II p. 290