Changing the variable until what is left is a standard form

What to assume they know

• The list of standard formulae, from module m01
• The two properties that split an integral across a sum and pull a constant out, from module m01
• The chain rule for derivatives, from Chapter 5 of Part I
• Differentials, and reading a derivative as a ratio that can be rearranged
• The Pythagorean identities and the definitions of the four derived trigonometric ratios, from Class XI
• The derivative of the inverse tangent, from Chapter 5 of Part I

What they should be able to do

• State the change-of-variable formula and identify each of its three pieces in a worked example
• Choose a substitution by looking for a function whose derivative already sits in the integrand
• Carry a substitution through completely, including the conversion of the differential, and return the answer to the original letter
• Recognise when one substitution leaves a second one to be made, and run both
• Derive the integral of the tangent and of the cotangent by substituting for the denominator
• Derive the integral of the secant and of the cosecant by the chapter's multiply-by-one device, and say what is being multiplied by what
• Manufacture a substitution where none is visible, by rewriting the integrand first
• Track a constant factor introduced by the substitution and account for it in the final answer
• Recognise, from the shape of an integrand, which of the standard substitutions it is asking for

Where it usually goes wrong

• "Substituting means replacing the variable and leaving everything else alone." The differential has to be converted too, and forgetting it is the single commonest error in the whole chapter. Show the conversion as its own step in section 4 and never let a later section skip it.
• "The substitution has to be for the innermost bracket." It has to be for something whose derivative is in the integrand. Example 6 (i) substitutes for the cosine while the bracket contains a square of the sine, and Example 5 (iii) substitutes twice, neither time for the innermost thing.
• "If I can't see a derivative in the integrand, substitution won't help." Example 6 (iii) puts one there by adding and subtracting; the secant and cosecant derivations put one there by multiplying above and below. The heuristic on Part II p. 236 finds the easy cases and the chapter quietly goes beyond it three times.
• "The answer can be left in the new variable." It cannot: the question was asked about the original letter, and an indefinite integral is a function, not a number. Section 8 exists for this. Note that the definite-integral section later offers a genuine exemption, and it is genuinely different — see module m03.
• "A constant that appears from the substitution can be absorbed into the constant of integration." Only an additive one can. A multiplicative one — the divisor in Example 5 (i) — scales the whole answer and must be carried.
• "The two printed forms of the tangent's integral are two different results." They are the same function written two ways, because the modulus of the secant is the reciprocal of the modulus of the cosine and the logarithm turns the reciprocal into a sign. Show the one-line reconciliation.
• "Multiplying above and below by that particular sum is a random trick." It is random the first time and structural afterwards: the sum chosen is exactly the one whose derivative is the numerator you end up with. Say what the target was and the choice stops being magic.
• "Substitution and the chain rule are different things." The method is the chain rule read from right to left, and saying so once makes every example predictable. The chapter never says it.

Questions to check understanding

• Identify a suitable substitution for a given integrand and justify the choice
• Carry out a substitution completely, including the differential, and return to the original variable
• Integrate a function requiring two successive substitutions — the form of Example 5 (iii)
• Derive the integral of the tangent or the cotangent from first principles
• Derive the integral of the secant or the cosecant using the chapter's device
• Integrate an odd power of a sine times a power of a cosine — the form of Example 6 (i)
• Integrate a reciprocal of one plus a trigonometric ratio, by manufacturing a derivative in the numerator — the form of Example 6 (iii)
• Choose the correct integral from four options — the form of Exercise 7.2 Q38 and Q39

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.

• §7.3's opening (Part II p. 235). It says inspection got the chapter this far and will not get it further, then lists three prominent methods: substitution, partial fractions, and parts. Note the mismatch before scripting section 1: the list has three items, and what actually follows is five blocks — §7.3.1 substitution, §7.3.2 trigonometric identities, §7.4 particular functions, §7.5 partial fractions, §7.6 parts. The chapter hedges with prominent among them, so the list is not claiming to be exhaustive.
• The change-of-variable derivation (§7.3.1, Part II pp. 235–236). Four lines: name the integral; set the old variable equal to a function of a new one; differentiate to convert the differential; substitute both. The result is an equation between two integrals in different letters. The chapter then says the hard part is guessing well, and gives the heuristic that follows.
• The heuristic (§7.3.1, Part II p. 236). Pick something whose own derivative is already one of the factors present. One sentence, and it is the only guidance the chapter offers about choosing; everything else is worked examples. Section 3 should state it, and section 9 should show the case it does not cover.
• Example 5 (Part II pp. 236–237), four parts. (i) The sine of a constant multiple: the substitution is the argument, and the constant reappears as a divisor. Verified: the answer is minus the cosine of the multiple, divided by the multiplier. (ii) Twice the variable times the sine of one more than its square: the derivative of the argument is exactly the outside factor, so nothing is left over. Verified: the answer is minus the cosine of one more than the square. (iii) The fourth power of the tangent of a square root, times the square of the secant of that root, all over the root. This is the double substitution — first the root, then the tangent — and the chapter also notes that the second one alone would have done. Verified: the answer is two fifths of the fifth power of the tangent of the root. (iv) The sine of an inverse tangent, over one plus a square. Verified: the answer is minus the cosine of the inverse tangent. Read the exponents and the inverse-function notation off the page image; the text layer flattens them.
• The four derived standard integrals (§7.3.1, Part II pp. 237–238), labelled (i) to (iv) and each with its own short derivation. (i) and (ii), the tangent and the cotangent: rewrite as a ratio of a sine and a cosine and substitute for the denominator. The tangent's answer is printed in two equivalent forms, minus the logarithm of the modulus of the cosine and the logarithm of the modulus of the secant, and the second is the one carried forward. Verified: the two differ by a sign inside a logarithm of a reciprocal, so they are the same function. (iii) and (iv), the secant and the cosecant: multiply the integrand above and below by a chosen sum, after which the numerator is the derivative of the denominator. This is the device of section 7 and it deserves its own attention, because nothing in the chapter's heuristic would ever suggest it. The chapter says outright that these four will be used later without further reference, which makes them part of the working vocabulary from here on.
• Example 6 (Part II pp. 238–240), three parts, each harder than Example 5. (i) An odd power of the sine times an even power of the cosine: peel one sine off, convert the rest by the Pythagorean identity, then substitute for the cosine. Verified: the answer is minus a third of the cube of the cosine plus a fifth of its fifth power. (ii) A sine over the sine of a shifted argument: substitute for the shifted argument, then expand the sine of a difference. Verified: the answer is the variable times the cosine of the shift, less the sine of the shift times the logarithm of the modulus of the sine of the shifted argument. Note that the chapter's own working produces a constant that absorbs two separate pieces, and it says so explicitly in the final line. (iii) The reciprocal of one plus the tangent: rewrite as a ratio of cosines, then add and subtract inside the numerator so that one half of it is the derivative of the denominator. This is section 9's example, and the trick is not an instance of the chapter's own heuristic — the derivative is not there until it is put there. Verified: the answer is half the variable plus half the logarithm of the modulus of the sum of the cosine and the sine.
• Exercise 7.2 (Part II pp. 240–241), thirty-nine items, of which the last two are multiple choice. Worth flagging for the script, by shape rather than by number: Q1, Q2, Q18, Q19, Q20, Q21, Q22, Q35, Q36 and Q37 all have the derivative visibly present and are the heuristic working as advertised; Q6, Q7, Q8, Q9 and Q16 are linear or quadratic substitutions inside a root; Q12, Q13 and Q14 need an index law first; Q23, Q26 and Q29 substitute for a whole composite; and Q25, Q32 and Q33 are the reciprocal-of-a-sum shape that Example 6 (iii) handles. Q10 is the one that is not what it looks like — it needs a rewrite before any substitution presents itself.
• Exercise 7.2, questions 38 and 39 (Part II p. 241). Two multiple-choice items. Verified by working added here: Q38's integrand is the derivative of a sum of a tenth power and an exponential with base ten divided by that same sum, so the answer is the logarithm of the sum, which is the fourth option. Q39's integrand is the reciprocal of a product of two squared ratios; rewriting it as the sum of the square of the secant and the square of the cosecant gives the tangent less the cotangent, which is the second option. The chapter prints no answers.
• The Summary's substitution block (Part II p. 289). It restates the method in two sentences and then reprints the four derived standard integrals as (i) to (iv). Read off the page image. It is a faithful condensation and can close the explanation.

Figures to have open

• A shape-to-substitution table for section 11, grouping the thirty-nine items of Exercise 7.2 into the families named above. The items are the chapter's own, from Part II pp. 240–241; the grouping is added here. Build it with the repo's DataTable component.
• A three-panel annotation of the change-of-variable equation for section 2, highlighting the old variable, the new variable and the converted differential in turn. Not in the book; the chapter prints the equation on Part II p. 236 with no annotation.
• A side-by-side reconciliation of the tangent's two printed answers for section 6. Both forms are the chapter's own, on Part II p. 237; the reconciliation is added here.
• 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", §7.3 Methods of Integration and its list of three methods, Part II p. 235
• §7.3.1 Integration by substitution, the change-of-variable derivation and the heuristic, Part II pp. 235–236
• Example 5, four parts, Part II pp. 236–237
• The four derived standard integrals, Part II pp. 237–238
• Example 6, three parts, Part II pp. 238–240
• Exercise 7.2, questions 1 to 39, Part II pp. 240–241
• Summary, the integration-by-substitution bullet, Part II p. 289