Integrating a product, and how the choice of first function decides whether it helps

What to assume they know

• The product rule for derivatives, from Chapter 5 of Part I
• The standard formulae, from module m01
• Substitution, from the first topic of this module, and the recognition that a product is not always a substitution in disguise
• The derivatives of the inverse sine, inverse cosine and inverse tangent, and of the natural logarithm
• Solving a linear equation in which the unknown appears on both sides
• The constant-factor and sum properties, from module m01

What they should be able to do

• Derive the by-parts formula by integrating the product rule, and identify each term in the result
• Name the two roles the chapter assigns to the two factors and say what each role requires of its factor
• Apply the formula both ways round on one integrand and say what the wrong choice produces
• State the chapter's two-tier rule of thumb for choosing the first function
• Integrate a function that is not visibly a product, by supplying a second factor
• Explain why no constant rides along when the second factor is integrated
• Handle the case where applying the formula twice returns the original integral, and solve for it
• State what the chapter claims about the integrand it says the method fails on, and distinguish that claim from a weaker one
• Decide, from a product, which factor should go first

Where it usually goes wrong

• "By parts is for products, so anything that is not a product is out of scope." Example 18 supplies a second factor of one and integrates a lone logarithm; Exercise 7.6 Q13 does the same for an inverse tangent. If the technique is presented as needing two visible factors, both items look impossible.
• "Either factor can go first; the answer is the same." The answer is the same when both routes terminate, and Example 17 shows a route that does not: the wrong choice raises the power and moves further from an answer. Show the failure, not just the success.
• "The rule of thumb is: powers first." It is two rules, and the second beats the first. Six of the fifteen items in this topic's slice of Exercise 7.6 are governed by the second tier.
• "A constant should be added to the second function's integral, to be safe." Remark (ii) carries one through and shows the extra terms annihilate. Adding it is not wrong, merely pointless, and the chapter says so in one word.
• "The method fails on the variable times a sine." It does not; that is Exercise 7.6 Q1. The integrand the chapter names is the square root of the variable times a sine, and the difference is one radical that the text layer loses. See the notes below.
• "When the integral reappears on the right, the method has failed." That is when it has succeeded: the equation can be solved. Students who have not met this pattern abandon Example 21 one line before the end.
• "Applying the formula twice always closes the loop." It closes for an exponential against a sine or cosine. Applied twice to a squared power against an exponential it lowers the power twice and terminates instead. Exercise 7.6 Q3 and Q14 are of the second kind.
• "Integration by parts is the product rule for integrals." There is no product rule for integrals. It is the product rule for derivatives, integrated and rearranged, and it converts one integral into a different one rather than evaluating anything by itself.

Questions to check understanding

• Derive the by-parts formula from the product rule
• Choose the first function for a stated product and justify the choice by the chapter's rule
• Integrate a product by parts — the form of Exercise 7.6 Q1 to Q9
• Integrate a lone logarithm or a lone inverse ratio by supplying a second factor — the form of Example 18 and of Exercise 7.6 Q13
• Integrate a product requiring the formula twice and then an equation solved — the form of Example 21 and of Exercise 7.6 Q10
• Show that an extra constant on the second factor's antiderivative changes nothing
• Explain what the chapter claims about the integrand it says the method does not reach, and what it does not claim

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 derivation (§7.6, Part II p. 259). Write the product rule for two differentiable functions, integrate both sides, and rearrange. The result is numbered as the section's first equation. The chapter then renames the two factors so the formula reads as an instruction rather than as an identity. Four lines.
• The formula in words (Part II p. 260). The chapter sets the whole instruction in a bold quoted sentence: take the first factor times whatever the second integrates to, then subtract the integral of that same antiderivative against the first factor's derivative. It uses differential coefficient where a modern reader expects derivative. Read off the page image. Show the words and the symbols together; students who learn only the symbols reverse the two roles under exam pressure.
• Example 17, both ways round (Part II p. 260). The variable times a cosine. Taken the sensible way, the answer falls out in two lines. Taken the other way round, the chapter carries the working far enough to show the resulting integral is worse than the one it started from — a higher power of the variable — and then says the choice matters. Verified: the correct answer is the variable times the sine, plus the cosine. This deliberate demonstration of the wrong choice is unusual and is the single most useful thing on the page; do not cut it.
• Remark (i) (Part II p. 260). Read this one off the page image and not off the text layer. The integrand the chapter names as beyond the method is a square root of the variable times a sine, and the sentence that follows repeats the same square root. The extraction drops both radicals, turning the example into the variable times a sine — which is Exercise 7.6 question 1, a routine by-parts item on the very next page. A brief or a script built from the text layer would therefore tell a student that the method fails on the first thing the exercise asks them to do. Confirmed on a 300 dots per inch the printed page of the source. Note also what the Remark actually claims: not that the method is unhelpful, but that no function has that integrand as its derivative — the same kind of statement as Remark (ii) on Part II p. 234.
• Remark (ii) (Part II pp. 260–261). Carrying an arbitrary constant through the antiderivative of the second factor and watching the extra terms cancel. The chapter calls the addition superfluous. Verified: the two extra terms are the constant times the first function and the integral of the constant times the first function's derivative, and they differ only in sign. Section 7 should run the cancellation rather than assert it.
• Remark (iii) (Part II p. 261). The rule of thumb, in two tiers: a power of the variable or a polynomial goes first; but if the other factor is an inverse trigonometric function or a logarithm, that one goes first instead. The second tier overrides the first, and Examples 18, 20 and Exercise 7.6 Q7 to Q11 are all cases where it does. The chapter offers no mnemonic and no ordering beyond these two sentences.
• Example 18 (Part II p. 261). A lone logarithm. There is no product until one is supplied: the logarithm is put in the first role and a constant factor of one is put in the second. Verified: the answer is the variable times the logarithm, less the variable. This is the example that makes the technique bigger than it looks, and it should have a section of its own.
• Example 19 (Part II p. 261). The variable times an exponential, the cleanest possible case. Verified: the answer is the exponential times one less than the variable, written by the chapter as two terms.
• Example 20 (Part II pp. 261–262). An inverse sine times the variable over the root of one less a square. The second factor's integral is found first, by its own substitution, before the by-parts formula is applied — a technique inside a technique, and the chapter flags an alternative route by substituting for the inverse sine. Verified by working added here: the second factor integrates to minus the root, and the final answer is the variable less the root times the inverse sine.
• Example 21 (Part II p. 262). An exponential times a sine. Applying the formula twice returns the original integral with a sign change, so the whole thing is a linear equation in the unknown integral. Verified: the answer is half the exponential times the difference of the sine and the cosine. The chapter notes that swapping the two roles works equally well, which is worth stating: this is the one case in the section where the choice does not matter, and that is exactly because the method is going to close the loop either way.
• Exercise 7.6, questions 1 to 15, 21 and 22 (Part II pp. 263–264). Grouped by which factor goes first — an added classification, not the chapter's: Q1, Q2, Q3, Q12 and Q14 put a power first, the rule's first tier; Q4, Q5, Q6 and Q15 put a logarithm first, the second tier; Q7, Q8, Q9, Q10, Q11 and Q13 put an inverse ratio first, the second tier again; and Q13 and Q22 are lone functions with a constant factor supplied alongside, the Example 18 shape. Q10 and Q21 both need the formula twice and then the answer solved for, like Example 21, and they are the two hardest items in the set. Q21 is worth noticing for a second reason: it is printed in the middle of a run of items that all belong to the next topic's shape, and it is not one of them. Questions 16 to 20 and the two multiple-choice items belong to the next topic and are cited there.
• The Summary (Part II p. 290). It prints the formula with the two factors subscripted rather than lettered, repeats the instruction in words, and closes with advice to take as second function whichever one integrates most easily — advice the section itself gives only through Remark (iii). Read off the page image.

Figures to have open

• A grouping table of Exercise 7.6 questions 1 to 15, 21 and 22 by which factor goes first, for section 11. The items are the chapter's own, from Part II pp. 263–264; the grouping is added here. Build it with the repo's DataTable component.
• A two-column working of Example 17 for section 4, the two role assignments run side by side to the point where one has terminated and the other has not. Both routes are the chapter's own, on Part II p. 260; the parallel layout is added here.
• A cancellation diagram for section 7, the two extra terms drawn and struck through. Not in the book; the chapter runs the algebra on Part II p. 261 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", §7.6 Integration by Parts, the derivation from the product rule, Part II p. 259
• The formula stated in words, and Example 17 worked both ways, Part II p. 260
• Remarks (i), (ii) and (iii), Part II pp. 260–261
• Examples 18 to 21, Part II pp. 261–262
• Exercise 7.6, questions 1 to 15, 21 and 22, Part II pp. 263–264
• Summary, Integration by parts, Part II p. 290