MathsISC Class 12 · Chapter 7

Integrals ISC Class 12: Methods, Properties and Solved Questions

Q: Is the proof of the fundamental theorem required?

No. The ISC syllabus states the Fundamental Theorem of Calculus without proof; you use it to evaluate definite integrals.

Q: Is definite integration as the limit of a sum required?

It is not listed in the ISC 2027 syllabus; definite integrals are evaluated with the fundamental theorem and the listed properties.

Q: Which special trigonometric integrals are named in the syllabus?

Among others: \int\frac{dx}{a\cos x + b\sin x}, \int\frac{dx}{a + b\cos x}, \int\frac{dx}{a + b\sin x}, \int\frac{dx}{a\cos x + b\sin x + c}, \int\frac{a\cos x + b\sin x}{c\cos x + d\sin x}dx, \int\frac{dx}{a\cos^2x + b\sin^2x + c}, \int\frac{x^2 \pm 1}{x^4 + 1}dx, \int\sqrt{\tan x}\,dx and \int\frac{dx}{x^4 + 1}.

Q: How do I know when to use a property of definite integrals?

If the integrand has no obvious antiderivative but the limits are 0 and a (or symmetric, or a and b), and replacing x by a − x (or a + b − x) produces a related expression, add the two forms.

Q: Do I always need + C?

Yes, for every indefinite integral; omitting it can cost a mark. Definite integrals do not need it.

Q: How many marks does integration carry?

It is part of Calculus (35 of 80 marks); expect integrals in Sections A, B, C and D.

Updated 14 October 2026 · Fact-checked

Integrals is Chapter 7 of ISC Class 12 Mathematics. It covers integration as anti-differentiation, standard and special integrals, integration by substitution, by parts and by partial fractions, trigonometric forms such as $\int\frac{dx}{a + b\cos x}$, the fundamental theorem of calculus (without proof), and the properties of definite integrals used to evaluate them.

What Integrals is about

Integration is the largest single topic in ISC Calculus (35 of 80 marks for the whole unit), and it appears in every section of the paper, from one-mark MCQs to 5-mark evaluations.

Indefinite integrals reverse differentiation: $\int f (x) d x = F (x) + C$ where F′ = f. You must know the standard results ( $x^{n}$ , $\frac{1}{x}$ , $e^{x}$ , trigonometric functions including tan, cot, sec and cosec) and the integrals of $sin^{2} x$ , $sin^{3} x$ , $cos^{4} x$ and similar powers.

Methods: (1) substitution, including the patterns $\int f^{'} (x) [f (x)]^{n} d x$ and $\int \frac{f ^{'} ( x )}{f ( x )} d x = lo g ∣ f (x) ∣$ ; (2) the special integrals $\int \frac{d x}{x ^{2} \pm a ^{2}}$ , $\int \frac{d x}{a ^{2} - x ^{2}}$ , $\int \frac{d x}{x ^{2} \pm a ^{2}}$ , $\int a^{2} \pm x^{2} d x$ and the quadratic forms reduced to them by completing the square, including $\int \frac{p x + q}{a x ^{2} + b x + c} d x$ ; (3) trigonometric forms such as $\int \frac{d x}{a + b c o s x}$ by the substitution $t = tan \frac{x}{2}$ , $\int tan x d x$ and $\int \frac{x ^{2} + 1}{x ^{4} + 1} d x$ ; (4) integration by parts, with the right choice of first function and the special case $\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x)$ ; (5) partial fractions for proper and improper rational functions, with linear, repeated linear and irreducible quadratic factors.

Definite integrals: the fundamental theorem (without proof), $\int_{a}^{b} f (x) d x = F (b) - F (a)$ , and the properties listed in the syllabus, especially $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ , $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ , the even/odd rule on [−a, a] and the $\int_{0}^{2 a}$ rule. Questions such as $\int_{0}^{π /2} \frac{s i n x}{s i n x + c o s x} d x$ cannot be done any other way.

Weightage: Integrals is part of the unit Calculus, which carries 35 of the 80 marks in the ISC Mathematics, shared with 4 other chapters. Questions on it can appear in any section of the paper.Source: CISCE ISC Mathematics syllabus, examination year 2027, distribution of marks

Key concepts

Standard integrals and integrals of trigonometric powers: Reverse each derivative: $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1}$ (n ≠ −1), $\int \frac{1}{x} d x = lo g ∣ x ∣$ , $\int e^{x} d x = e^{x}$ , $\int a^{x} d x = \frac{a ^{x}}{l o g a}$ , $\int sin x = - cos x$ , $\int cos x = sin x$ , $\int sec^{2} x = tan x$ , $\int cosec^{2} x = - cot x$ , $\int tan x = lo g ∣ sec x ∣$ , $\int cot x = lo g ∣ sin x ∣$ , $\int sec x = lo g ∣ sec x + tan x ∣$ , $\int cosec x = lo g ∣ cosec x - cot x ∣$ . For a linear inside, $\int f (a x + b) d x = \frac{1}{a} F (a x + b)$ . Powers: $sin^{2} x = \frac{1 - c o s 2 x}{2}$ ; $sin^{3} x = \frac{3 s i n x - s i n 3 x}{4}$ ; $cos^{4} x$ via $(cos^{2} x)^{2}$ . Always add + C.; Worked example
Evaluate (a) $\int sin^{2} x d x$ , (b) $\int \frac{5}{3 x + 1} d x$ .
Try it first, then show the solutionSolution
(a) $\int \frac{1 - c o s 2 x}{2} d x = \frac{x}{2} - \frac{s i n 2 x}{4} + C$
(b) $5 \int (3 x + 1)^{- 1/2} d x = 5 \cdot \frac{( 3 x + 1 ) ^{1/2}}{\frac{1}{2} \cdot 3} + C = \frac{10}{3} 3 x + 1 + C$
Answer: (a) $\frac{x}{2} - \frac{s i n 2 x}{4} + C$ (b) $\frac{10}{3} 3 x + 1 + C$
Integration by substitution: If the integrand contains a function and (a multiple of) its derivative, substitute t = that function. Key patterns: $\int f^{'} (x) [f (x)]^{n} d x = \frac{[ f ( x ) ] ^{n + 1}}{n + 1}$ ; $\int \frac{f ^{'} ( x )}{f ( x )} d x = lo g ∣ f (x) ∣$ ; $\int e^{f (x)} f^{'} (x) d x = e^{f (x)}$ . Change every x (and dx) into t, integrate, and substitute back. In definite integrals, change the limits too.; Worked example
Evaluate (a) $\int \frac{e ^{t a n^{- 1} x}}{1 + x ^{2}} d x$ and (b) $\int \frac{2 x + 3}{x ^{2} + 3 x + 5} d x$ .
Try it first, then show the solutionSolution
(a) t = $tan^{- 1} x$ , dt = $\frac{d x}{1 + x ^{2}}$ : $\int e^{t} d t = e^{t a n^{- 1} x} + C$ .
(b) Numerator is the derivative of the denominator: $lo g ∣ x^{2} + 3 x + 5∣ + C$ .
Answer: (a) $e^{t a n^{- 1} x} + C$ (b) $lo g (x^{2} + 3 x + 5) + C$
Special integrals and completing the square: $\int \frac{d x}{x ^{2} + a ^{2}} = \frac{1}{a} tan^{- 1} \frac{x}{a}$ ; $\int \frac{d x}{x ^{2} - a ^{2}} = \frac{1}{2 a} lo g \frac{x - a}{x + a}$ ; $\int \frac{d x}{a ^{2} - x ^{2}} = \frac{1}{2 a} lo g \frac{a + x}{a - x}$ ; $\int \frac{d x}{a ^{2} - x ^{2}} = sin^{- 1} \frac{x}{a}$ ; $\int \frac{d x}{x ^{2} \pm a ^{2}} = lo g ∣ x + x^{2} \pm a^{2} ∣$ . For $a x^{2} + b x + c$ under the integral, complete the square. For $\int \frac{p x + q}{a x ^{2} + b x + c} d x$ , write px + q = A(derivative of denominator) + B; the first part gives a log, the second a special integral.; Worked example
Evaluate $\int \frac{x + 3}{x ^{2} + 2 x + 5} d x$ .
Try it first, then show the solutionSolution
x + 3 = ½(2x + 2) + 2.
$\frac{1}{2} \int \frac{2 x + 2}{x ^{2} + 2 x + 5} d x = \frac{1}{2} lo g (x^{2} + 2 x + 5)$ .
$2 \int \frac{d x}{( x + 1 ) ^{2} + 2 ^{2}} = 2 \cdot \frac{1}{2} tan^{- 1} \frac{x + 1}{2} = tan^{- 1} \frac{x + 1}{2}$ .
Answer: $\frac{1}{2} lo g (x^{2} + 2 x + 5) + tan^{- 1} \frac{x + 1}{2} + C$
Trigonometric forms: a + b cos x and similar: For $\int \frac{d x}{a + b c o s x}$ , $\int \frac{d x}{a + b s i n x}$ and $\int \frac{d x}{a c o s x + b s i n x + c}$ , put $t = tan \frac{x}{2}$ : $cos x = \frac{1 - t ^{2}}{1 + t ^{2}}$ , $sin x = \frac{2 t}{1 + t ^{2}}$ , $d x = \frac{2 d t}{1 + t ^{2}}$ ; the integral becomes a special rational integral in t. For $\int \frac{d x}{a c o s ^{2} x + b s i n ^{2} x + c}$ , divide by $cos^{2} x$ and put t = tan x. For $\int \frac{a c o s x + b s i n x}{c c o s x + d s i n x} d x$ , write numerator = A(denominator) + B(derivative of denominator).; Worked example
Evaluate $\int \frac{d x}{5 + 4 c o s x}$ .
Try it first, then show the solutionSolution
t = tan(x/2): $5 + 4 \frac{1 - t ^{2}}{1 + t ^{2}} = \frac{9 + t ^{2}}{1 + t ^{2}}$ ; $d x = \frac{2 d t}{1 + t ^{2}}$ .
$\int \frac{2 d t}{9 + t ^{2}} = \frac{2}{3} tan^{- 1} \frac{t}{3}$ .
Answer: $\frac{2}{3} tan^{- 1} (\frac{1}{3} tan \frac{x}{2}) + C$
Integration by parts: $\int u v d x = u \int v d x - \int (u^{'} \int v d x) d x$ . Choose u (the first function) by ILATE: Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential, taking the earlier one as u. For $\int lo g x d x$ or $\int tan^{- 1} x d x$ , take 1 as the second function. Repeat parts for $x^{2} e^{x}$ or $x^{2} cos x$ . Special result: $\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C$ .; Worked example
Evaluate $\int x^{2} e^{x} d x$ and $\int lo g x d x$ .
Try it first, then show the solutionSolution
$\int x^{2} e^{x} d x = x^{2} e^{x} - \int 2 x e^{x} d x = x^{2} e^{x} - 2 (x e^{x} - e^{x}) = e^{x} (x^{2} - 2 x + 2) + C$ .
$\int lo g x \cdot 1 d x = x lo g x - \int \frac{1}{x} \cdot x d x = x lo g x - x + C$ .
Answer: $e^{x} (x^{2} - 2 x + 2) + C$ ; $x lo g x - x + C$
Partial fractions: For a proper rational function $\frac{f ( x )}{g ( x )}$ (degree f < degree g), split the denominator: (x − a)(x − b) gives $\frac{A}{x - a} + \frac{B}{x - b}$ ; a repeated factor $(x - a)^{2}$ gives $\frac{A}{x - a} + \frac{B}{( x - a ) ^{2}}$ ; an irreducible quadratic $x^{2} + p x + q$ gives $\frac{A x + B}{x ^{2} + p x + q}$ . Find constants by substituting convenient x values and comparing coefficients. For an improper fraction (degree f ≥ degree g), divide first: e.g. $\frac{x ^{2} + 1}{x ^{2} - 5 x + 6} = 1 + \frac{5 x - 5}{x ^{2} - 5 x + 6}$ .; Worked example
Evaluate $\int \frac{x + 2}{( x - 3 ) ( x + 1 )} d x$ .
Try it first, then show the solutionSolution
x + 2 = A(x + 1) + B(x − 3). x = 3: 5 = 4A, A = 5/4. x = −1: 1 = −4B, B = −1/4.
$\int (\frac{5/4}{x - 3} - \frac{1/4}{x + 1}) d x$
Answer: $\frac{5}{4} lo g ∣ x - 3∣ - \frac{1}{4} lo g ∣ x + 1∣ + C$
Definite integrals and the fundamental theorem: Fundamental theorem of calculus (without proof): if F′ = f on [a, b], then $\int_{a}^{b} f (x) d x = F (b) - F (a)$ ; no constant is needed. With modulus functions, split the interval where the expression inside changes sign. With substitution, change the limits along with the variable.; Worked example
Evaluate (a) $\int_{0}^{π /2} sin^{2} x d x$ and (b) $\int_{0}^{2} ∣ x - 1∣ d x$ .
Try it first, then show the solutionSolution
(a) $[\frac{x}{2} - \frac{s i n 2 x}{4}]_{0}^{π /2} = \frac{π}{4}$ .
(b) $\int_{0}^{1} (1 - x) d x + \int_{1}^{2} (x - 1) d x = \frac{1}{2} + \frac{1}{2} = 1$ .
Answer: (a) $\frac{π}{4}$ (b) 1
Properties of definite integrals: (1) $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (t) d t$ ; (2) $\int_{a}^{b} f = - \int_{b}^{a} f$ ; (3) $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; (4) $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ ; (5) $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ ; (6) $\int_{0}^{2 a} f = 2 \int_{0}^{a} f$ if f(2a − x) = f(x), and 0 if f(2a − x) = −f(x); (7) $\int_{- a}^{a} f = 2 \int_{0}^{a} f$ if f is even, 0 if f is odd. Typical trick: call the integral I, apply property (5) or (4) to get a second form of I, add the two forms so the awkward part cancels, and solve for I.; Worked example
Evaluate $I = \int_{0}^{π /2} \frac{s i n x}{s i n x + c o s x} d x$ .
Try it first, then show the solutionSolution
Property (5) with a = π/2 swaps sin and cos: $I = \int_{0}^{π /2} \frac{c o s x}{c o s x + s i n x} d x$ .
Adding: $2 I = \int_{0}^{π /2} 1 d x = \frac{π}{2}$ , so $I = \frac{π}{4}$ .
Answer: $\frac{π}{4}$

Maths in real life: Integrals

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Definite integral as total change

Suppose a car starting from rest has velocity v(t) = 3t² + 2t m s⁻¹ during the first few seconds.

How far does it travel in the first 3 seconds, and in the third second alone?

Try it first, then show the solutionSolution

$\int_{0}^{3} (3 t^{2} + 2 t) d t = [t^{3} + t^{2}]_{0}^{3} = 27 + 9 = 36$ m
Third second: $[t^{3} + t^{2}]_{2}^{3} = 36 - 12 = 24$ m

Answer: 36 m in 3 s; 24 m during the third second

Memory hook: Integrate a rate to get the total.

Accumulating a varying rate

Imagine water flowing into a school's storage tank at r(t) = 20 − t litres per minute for 10 minutes as the supply pressure falls.

How much water enters the tank?

Try it first, then show the solutionSolution

$\int_{0}^{10} (20 - t) d t = [20 t - \frac{t ^{2}}{2}]_{0}^{10}$
= 200 − 50 = 150

Answer: 150 litres

Memory hook: Area under a rate graph = amount gathered.

Integrating an exponential rate

Suppose a phone charges at a rate of $2 e^{- 0.1 t}$ per cent per minute, slowing as the battery fills.

By how much does the charge rise in the first 10 minutes?

Try it first, then show the solutionSolution

$\int_{0}^{10} 2 e^{- 0.1 t} d t = [- 20 e^{- 0.1 t}]_{0}^{10} = 20 (1 - e^{- 1})$
$= 20 \times 0.632 = 12.6$

Answer: About 12.6 percentage points

Memory hook: $\int e^{k t} d t = \frac{1}{k} e^{k t}$ : remember to divide by k.

Topics in this chapter

Integrals formula sheet

Power and log: $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C (n \neq = - 1), \int \frac{d x}{x} = lo g ∣ x ∣ + C$
Trigonometric: $\int tan x d x = lo g ∣ sec x ∣, \int cot x d x = lo g ∣ sin x ∣, \int sec x d x = lo g ∣ sec x + tan x ∣, \int cosec x d x = lo g ∣ cosec x - cot x ∣$; + C each.
Special integrals I: $\int \frac{d x}{x ^{2} + a ^{2}} = \frac{1}{a} tan^{- 1} \frac{x}{a}, \int \frac{d x}{x ^{2} - a ^{2}} = \frac{1}{2 a} lo g \frac{x - a}{x + a}, \int \frac{d x}{a ^{2} - x ^{2}} = \frac{1}{2 a} lo g \frac{a + x}{a - x}$
Special integrals II: $\int \frac{d x}{a ^{2} - x ^{2}} = sin^{- 1} \frac{x}{a}, \int \frac{d x}{x ^{2} \pm a ^{2}} = lo g ∣ x + x^{2} \pm a^{2} ∣$
Square-root integrals: $\int a^{2} - x^{2} d x = \frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} \frac{x}{a}; \int x^{2} \pm a^{2} d x = \frac{x}{2} x^{2} \pm a^{2} \pm \frac{a ^{2}}{2} lo g ∣ x + x^{2} \pm a^{2} ∣$; + C.
Integration by parts: $\int uv d x = u \int v d x - \int (\frac{d u}{d x} \int v d x) d x$; ILATE for choosing u.
e^x rule: $\int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C$
Half-angle substitution: $t = tan \frac{x}{2} : cos x = \frac{1 - t ^{2}}{1 + t ^{2}}, sin x = \frac{2 t}{1 + t ^{2}}, d x = \frac{2 d t}{1 + t ^{2}}$; For a + b cos x, a + b sin x.
Fundamental theorem: $\int_{a}^{b} f (x) d x = F (b) - F (a)$; Without proof.
Key properties: $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x; \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$; Add the two forms to simplify.
Even, odd and 2a rules: $\int_{- a}^{a} f = 2 \int_{0}^{a} f (even), 0 (odd); \int_{0}^{2 a} f = 2 \int_{0}^{a} f if f (2 a - x) = f (x)$; 0 if f(2a − x) = −f(x).

How the board asks Integrals

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
One-mark evaluations	Standard definite integrals, $\int e^{x} (f + f^{'})$ , modulus integrals, find-the-error and 'whose answer is correct' items.	Recognise the standard form; remember $\frac{1}{a}$ for a linear inside.
Substitution and special forms (2-3 marks)	$\int \frac{p x + q}{a x ^{2} + b x + c}$ , $\int \frac{d x}{a x ^{2} + b x + c}$ , $\int \frac{c o s x}{a c o s x + b}$ .	Split the numerator; complete the square; reduce to a table form.
By parts	$\int x^{2} cos x$ , $\int (lo g x)^{2}$ , $\int e^{x} sin x$ , $\int tan^{- 1} x$ .	ILATE; write the formula before applying; repeat if needed.
Partial fractions (3-5 marks)	Linear, repeated linear and quadratic factors; improper fractions.	Divide first if improper; set up the right template; find constants by substitution.
Properties of definite integrals (3-5 marks)	Evaluate integrals like $\int_{0}^{π} \frac{x s i n x}{1 + c o s ^{2} x} d x$ or $\int_{0}^{2 π} \frac{d x}{1 + e ^{s i n x}}$ .	Name I, apply $f (a - x)$ or $f (a + b - x)$ , add, solve.
Real-life accumulation	Distance from velocity, total quantity from a rate.	Total change = definite integral of the rate over the time interval.

Tricks to remember Integrals

Short memory aids. Say each one aloud once, then use it on a question the same day.

ILATE
Inverse trig, Log, Algebraic, Trig, Exponential: the earlier one is the first function u.
Use it when: Choosing u in integration by parts.
Derivative on top means log
If the numerator is the derivative of the denominator, the answer is log|denominator|.
Use it when: Spotting $\int \frac{f ^{'}}{f}$ .
Plus-a-squared gives tan⁻¹
x² + a² → tan⁻¹; a² − x² under a root → sin⁻¹; x² ± a² under a root → log.
Use it when: Choosing the special integral.
Name it, flip it, add it
For property questions: call it I, replace x by (a − x), add the two versions.
Use it when: Definite-integral properties.
Divide by a for (ax + b)
A linear inside the function divides the result by its coefficient.
Use it when: Avoiding the missing ½ examiners report.

Integrals: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: A standard definite integral · 1 mark · MCQBoard level

The value of $\int_{0}^{1} \frac{d x}{1 + x ^{2}}$ is:
(a) $\frac{π}{2}$
(b) $\frac{π}{3}$
(c) $\frac{π}{4}$
(d) $\frac{π}{6}$

Try it first, then show the solutionSolution

$[tan^{- 1} x]_{0}^{1} = \frac{π}{4} - 0$ .

Answer: (c)

Example 2: An odd integrand · 1 mark · Assertion-reasonBoard level

Assertion: $\int_{- π /2}^{π /2} x sin^{2} x d x = 0$ .
Reason: If f is an odd function, $\int_{- a}^{a} f (x) d x = 0$ .
(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Try it first, then show the solutionSolution

f(−x) = (−x)sin²(−x) = −x sin²x, so f is odd and the integral over a symmetric interval vanishes; the reason gives exactly this.

Answer: (a)

Example 3: A linear inside a power · 2 marks · Short answerHarder board level

Evaluate $\int \frac{5}{3 x + 1} d x$ . [2]

Try it first, then show the solutionSolution

$5 \int (3 x + 1)^{- 1/2} d x = 5 \frac{( 3 x + 1 ) ^{1/2}}{( 1/2 ) \times 3}$
$= \frac{10}{3} 3 x + 1 + C$

Answer: $\frac{10}{3} 3 x + 1 + C$

Example 4: Parts twice · 3 marks · Short answerHarder board level

Evaluate $\int x^{2} e^{x} d x$ . [3]

Try it first, then show the solutionSolution

u = x² (algebraic before exponential): $x^{2} e^{x} - \int 2 x e^{x} d x$ .
Again by parts: $\int x e^{x} d x = x e^{x} - e^{x}$ .
$x^{2} e^{x} - 2 x e^{x} + 2 e^{x} + C = e^{x} (x^{2} - 2 x + 2) + C$ .

Answer: $e^{x} (x^{2} - 2 x + 2) + C$

Example 5: Linear over quadratic · 3 marks · Short answerHarder board level

Evaluate $\int \frac{x + 3}{x ^{2} + 2 x + 5} d x$ . [3]

Try it first, then show the solutionSolution

Write x + 3 = ½(2x + 2) + 2.
First part: $\frac{1}{2} lo g (x^{2} + 2 x + 5)$ .
Second part: $2 \int \frac{d x}{( x + 1 ) ^{2} + 4} = tan^{- 1} \frac{x + 1}{2}$ .

Answer: $\frac{1}{2} lo g (x^{2} + 2 x + 5) + tan^{- 1} \frac{x + 1}{2} + C$

Example 6: Partial fractions with a repeated factor · 5 marks · Long answerHarder board level

(a) Resolve $\frac{3 x + 1}{( x - 1 ) ^{2} ( x + 2 )}$ into partial fractions. [3]
(b) Hence evaluate $\int \frac{3 x + 1}{( x - 1 ) ^{2} ( x + 2 )} d x$ . [2]

Try it first, then show the solutionSolution

(a) $\frac{3 x + 1}{( x - 1 ) ^{2} ( x + 2 )} = \frac{A}{x - 1} + \frac{B}{( x - 1 ) ^{2}} + \frac{C}{x + 2}$ , so 3x + 1 = A(x − 1)(x + 2) + B(x + 2) + C(x − 1)².
x = 1: 4 = 3B, B = 4/3. x = −2: −5 = 9C, C = −5/9.
Coefficient of x²: 0 = A + C, so A = 5/9.
(b) $\int = \frac{5}{9} lo g ∣ x - 1∣ + \frac{4}{3} \int (x - 1)^{- 2} d x - \frac{5}{9} lo g ∣ x + 2∣$
$= \frac{5}{9} lo g \frac{x - 1}{x + 2} - \frac{4}{3 ( x - 1 )} + C$

Answer: $\frac{5}{9} lo g \frac{x - 1}{x + 2} - \frac{4}{3 ( x - 1 )} + C$

Example 7: Using the f(a − x) property · 3 marks · Short answerHOTS / competency

Evaluate $\int_{0}^{π /2} \frac{s i n x}{s i n x + c o s x} d x$ . [3]

Try it first, then show the solutionSolution

Let I be the integral. Using $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ with a = π/2: $I = \int_{0}^{π /2} \frac{c o s x}{c o s x + s i n x} d x$ .
Add: $2 I = \int_{0}^{π /2} d x = \frac{π}{2}$ .
$I = \frac{π}{4}$ .

Answer: $\frac{π}{4}$

Example 8: The e^x[f + f′] pattern · 2 marks · Short answerHOTS / competency

Evaluate $\int e^{x} (\frac{1}{x} - \frac{1}{x ^{2}}) d x$ . [2]

Try it first, then show the solutionSolution

With f(x) = 1/x, f′(x) = −1/x², so the integrand is $e^{x} [f (x) + f^{'} (x)]$ .
Result: $e^{x} f (x) + C = \frac{e ^{x}}{x} + C$ .

Answer: $\frac{e ^{x}}{x} + C$

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Distance from a velocity record

Suppose a car's speedometer data show that, starting from rest, its velocity is v(t) = 3t² + 2t m s⁻¹ for the first 4 seconds.

(i) · 1 mark
Write the distance travelled in the first T seconds as a definite integral.
Answer
Answer: $\int_{0}^{T} (3 t^{2} + 2 t) d t$
Distance = integral of speed over time (v ≥ 0 here).
(ii) · 1 mark
Find the distance in the first 3 s.
Answer
Answer: 36 m
$[t^{3} + t^{2}]_{0}^{3} = 36$
(iii) · 1 mark
Find the distance covered during the third second.
Answer
Answer: 24 m
$[t^{3} + t^{2}]_{2}^{3} = 36 - 12$
(iv) · 2 marks
Find the average velocity over the first 4 s.
Answer
Answer: 20 m s⁻¹
Distance $= [t^{3} + t^{2}]_{0}^{4} = 64 + 16 = 80$ m; average = 80/4 = 20 m s⁻¹.

Two integrals by one trick

Suppose a teacher asks a class to evaluate $I = \int_{0}^{π /2} \frac{d x}{1 + t a n x}$ and $J = \int_{0}^{π /2} lo g (tan x) d x$ without finding antiderivatives.

(i) · 1 mark
Write $\frac{1}{1 + t a n x}$ in terms of sin and cos.
Answer
Answer: $\frac{c o s x}{c o s x + s i n x}$
Multiply numerator and denominator by $cos x$ .
(ii) · 2 marks
Use $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$ to find I.
Answer
Answer: $\frac{π}{4}$
Replacing x by π/2 − x swaps sin and cos; adding, 2I = π/2.
(iii) · 1 mark
Show that J = 0.
Answer
Answer: J = 0
Replacing x by π/2 − x gives $lo g (cot x) = - lo g (tan x)$ , so J = −J.
(iv) · 1 mark
Which property was used in both parts?
Answer
Answer: $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x$
With a = π/2, it exchanges sin and cos (and tan with cot).

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Evaluate $\int \frac{5}{3 x + 1} d x$ . (2 marks)

Step	Marks
Writes as $5 \int (3 x + 1)^{- 1/2} d x$ and integrates the power	1
Divides by 3 for the linear inside: $\frac{10}{3} 3 x + 1 + C$	1

Evaluate $\int \frac{x + 2}{( x - 3 ) ( x + 1 )} d x$ . (3 marks)

Step	Marks
Correct partial-fraction form	1
A = 5/4, B = −1/4	1
$\frac{5}{4} lo g ∣ x - 3∣ - \frac{1}{4} lo g ∣ x + 1∣ + C$	1

Evaluate $\int_{0}^{π} \frac{x s i n x}{1 + c o s ^{2} x} d x$ . (5 marks)

Step	Marks
Names I and applies $f (π - x)$ : $I = \int_{0}^{π} \frac{( π - x ) s i n x}{1 + c o s ^{2} x} d x$	1
Adds: $2 I = π \int_{0}^{π} \frac{s i n x}{1 + c o s ^{2} x} d x$	1
Substitutes t = cos x with new limits 1 to −1	1
$2 I = π \int_{- 1}^{1} \frac{d t}{1 + t ^{2}} = π [tan^{- 1} t]_{- 1}^{1} = \frac{π ^{2}}{2}$	1
$I = \frac{π ^{2}}{4}$	1

Where students lose marks in Integrals

The slips examiners see most often, and what to write instead.

Mistake: Wrong first function in integration by parts
CISCE examiners' comments, 2024 and 2025: in $\int x^{2} cos x d x$ and $\int (lo g x)^{2} d x$ , candidates chose the first and second functions wrongly and did not write the standard form of the parts formula.
Wrong working
Takes cos x as u in $\int x^{2} cos x d x$ , making the integral harder each time.
Right working
u = x² (algebraic before trigonometric); for $(lo g x)^{2}$ take u = $(lo g x)^{2}$ and v = 1.
Do this: Write the formula first and choose u by ILATE.
Usually costs: 2-3 marks.
Mistake: Not splitting px + q correctly
CISCE examiners' comments, 2024: in $\int \frac{x + 7}{x ^{2} + 4 x + 7} d x$ candidates could not split the integral into two parts and could not bring the second part to a special form.
Wrong working
Integrates numerator and denominator separately.
Right working
x + 7 = ½(2x + 4) + 5; then a log term plus $5 \int \frac{d x}{( x + 2 ) ^{2} + 3}$ .
Do this: Match the derivative of the denominator first, then complete the square.
Usually costs: 2 marks.
Mistake: Missing the property in a definite integral
CISCE examiners' comments, 2024 and 2025: candidates failed to recognise that integrals like $\int_{0}^{2 π} \frac{d x}{1 + e ^{s i n x}}$ or $\int_{π /4}^{3 π /4} \frac{x d x}{1 + s i n x}$ need a property, used an incorrect property or shifted π wrongly.
Wrong working
Tries to find an antiderivative of $\frac{1}{1 + e ^{s i n x}}$ .
Right working
Apply $f (a + b - x)$ (here $sin (2 π - x) = - sin x$ ), add the two forms: $2 I = \int_{0}^{2 π} 1 d x$ , I = π.
Do this: If no antiderivative appears, look for a property.
Usually costs: Up to the whole question.
Mistake: Forgetting the factor from a linear inside
CISCE examiners' comments, 2025: in $\int \frac{5}{2 x + 7} d x$ candidates did not multiply by ½ after integrating $x^{- 1/2}$ .
Wrong working
$10 2 x + 7 + C$
Right working
$5 \cdot \frac{( 2 x + 7 ) ^{1/2}}{( 1/2 ) \cdot 2} = 5 2 x + 7 + C$
Do this: Differentiate your answer mentally to check.
Usually costs: 1 mark.
Mistake: Wrong partial-fraction template
CISCE examiners' comments, 2025: for $\frac{x ^{2}}{( x - 1 ) ^{2} ( x ^{2} + 1 )}$ candidates could not split into partial fractions correctly; in 2025 others did not reduce $\frac{c o s x}{3 c o s x - 5}$ to the standard form $\frac{d x}{a \pm b c o s x}$ .
Wrong working
$\frac{A}{( x - 1 ) ^{2}} + \frac{B}{x ^{2} + 1}$
Right working
$\frac{A}{x - 1} + \frac{B}{( x - 1 ) ^{2}} + \frac{C x + D}{x ^{2} + 1}$
Do this: A repeated factor needs every power; a quadratic factor needs Cx + D.
Usually costs: 2-3 marks.

Practice questions on Integrals

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
$\int sec^{2} x d x$ equals:
(a) tan x + C
(b) sec x tan x + C
(c) −cot x + C
(d) sec x + C
Hint
Which function has derivative sec²x?
Answer
(a)
Q2 · 1 mark · MCQBoard level
$\int tan x d x$ equals:
(a) sec²x + C
(b) log|cos x| + C
(c) log|sec x| + C
(d) log|sin x| + C
Hint
tan x = sin x/cos x.
Answer
(c)
Q3 · 1 mark · MCQBoard level
The value of $\int_{0}^{2} ∣ x - 1∣ d x$ is:
(a) 0
(b) 1
(c) 2
(d) ½
Hint
Two triangles.
Answer
(b) 1
Q4 · 1 mark · Very short answerBoard level
Evaluate $\int e^{3 x} d x$ .
Hint
Divide by 3.
Answer
$\frac{1}{3} e^{3 x} + C$
Q5 · 2 marks · Short answerBoard level
Evaluate $\int sin^{2} x d x$ .
Hint
Double-angle identity.
Answer
$\frac{x}{2} - \frac{s i n 2 x}{4} + C$
Q6 · 2 marks · Short answerHarder board level
Evaluate $\int \frac{d x}{x ^{2} + 4 x + 13}$ .
Hint
$(x + 2)^{2} + 9$ .
Answer
$\frac{1}{3} tan^{- 1} \frac{x + 2}{3} + C$
Q7 · 2 marks · Short answerHarder board level
Evaluate $\int lo g x d x$ .
Hint
Take 1 as the second function.
Answer
$x lo g x - x + C$
Q8 · 2 marks · Short answerHOTS / competency
Show that $\int_{0}^{π /2} lo g (tan x) d x = 0$ .
Hint
Replace x by π/2 − x.
Answer
I = $\int_{0}^{π /2} lo g (cot x) d x = - \int_{0}^{π /2} lo g (tan x) d x = - I$ , so I = 0.
Q9 · 3 marks · Short answerHarder board level
Evaluate $\int \frac{x + 2}{( x - 3 ) ( x + 1 )} d x$ .
Hint
A = 5/4.
Answer
$\frac{5}{4} lo g ∣ x - 3∣ - \frac{1}{4} lo g ∣ x + 1∣ + C$
Q10 · 3 marks · Short answerHarder board level
Evaluate $\int \frac{d x}{5 + 4 c o s x}$ .
Hint
t = tan(x/2).
Answer
$\frac{2}{3} tan^{- 1} (\frac{1}{3} tan \frac{x}{2}) + C$
Q11 · 3 marks · Short answerHarder board level
Evaluate $\int_{0}^{2} 4 - x^{2} d x$ and interpret the answer geometrically.
Hint
Use the √(a² − x²) formula.
Answer
$[\frac{x}{2} 4 - x^{2} + 2 sin^{- 1} \frac{x}{2}]_{0}^{2} = 2 \cdot \frac{π}{2} = π$ : the area of a quarter circle of radius 2.
Q12 · 5 marks · Long answerHarder board level
Evaluate $\int_{0}^{π} \frac{x s i n x}{1 + c o s ^{2} x} d x$ .
Hint
f(π − x), then t = cos x.
Answer
Applying f(π − x) and adding: $2 I = π \int_{0}^{π} \frac{s i n x}{1 + c o s ^{2} x} d x = π \int_{- 1}^{1} \frac{d t}{1 + t ^{2}} = π \cdot \frac{π}{2}$ , so $I = \frac{π ^{2}}{4}$ .
Q13 · 2 marks · Short answerHarder board level
Evaluate $\int_{0}^{π /2} sin^{2} x d x$ .
Hint
Or use the property with cos²x.
Answer
$\frac{π}{4}$

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. $\int sec x d x$ = ?Show answer
$lo g ∣ sec x + tan x ∣ + C$ .
2. $\int \frac{d x}{x ^{2} + a ^{2}}$ = ?Show answer
$\frac{1}{a} tan^{- 1} \frac{x}{a} + C$ .
3. $\int \frac{d x}{a ^{2} - x ^{2}}$ = ?Show answer
$sin^{- 1} \frac{x}{a} + C$ .
4.State the integration by parts formula.Show answer
$\int uv d x = u \int v d x - \int (u^{'} \int v d x) d x$ .
5. $\int e^{x} [f (x) + f^{'} (x)] d x$ = ?Show answer
$e^{x} f (x) + C$ .
6.Partial fraction for a repeated factor $(x - a)^{2}$ ?Show answer
$\frac{A}{x - a} + \frac{B}{( x - a ) ^{2}}$ .
7. $\int_{0}^{a} f (x) d x$ = ?Show answer
$\int_{0}^{a} f (a - x) d x$ .
8. $\int_{- a}^{a} f (x) d x$ for odd f?Show answer
0.

Best order to study it

Learn the standard table and trigonometric power integrals.
Practise substitution patterns ( $f^{'} / f$ , $f^{'} f^{n}$ , $e^{f} f^{'}$ ).
Learn the special integrals and completing the square, then $\frac{p x + q}{q u a d r a t i c}$ forms.
Practise trigonometric forms with t = tan(x/2).
Master integration by parts and the $e^{x} [f + f^{'}]$ rule.
Practise partial fractions of all types.
Finish with definite integrals and every property in the syllabus.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you write all standard integrals, including sec x and cosec x?
Can you integrate sin²x, sin³x and cos⁴x?
Can you recognise $f^{'} / f$ and $f^{'} f^{n}$ patterns?
Can you complete the square and choose the right special integral?
Can you split $\frac{p x + q}{a x ^{2} + b x + c}$ ?
Can you choose u by ILATE and repeat parts?
Can you set up partial fractions for repeated and quadratic factors?
Can you use $f (a - x)$ and $f (a + b - x)$ to evaluate definite integrals?

Integrals: frequently asked questions

Is the proof of the fundamental theorem required?

No. The ISC syllabus states the Fundamental Theorem of Calculus without proof; you use it to evaluate definite integrals.

Is definite integration as the limit of a sum required?

It is not listed in the ISC 2027 syllabus; definite integrals are evaluated with the fundamental theorem and the listed properties.

Which special trigonometric integrals are named in the syllabus?

Among others: \int\frac{dx}{a\cos x + b\sin x}, \int\frac{dx}{a + b\cos x}, \int\frac{dx}{a + b\sin x}, \int\frac{dx}{a\cos x + b\sin x + c}, \int\frac{a\cos x + b\sin x}{c\cos x + d\sin x}dx, \int\frac{dx}{a\cos^2x + b\sin^2x + c}, \int\frac{x^2 \pm 1}{x^4 + 1}dx, \int\sqrt{\tan x}\,dx and \int\frac{dx}{x^4 + 1}.

How do I know when to use a property of definite integrals?

If the integrand has no obvious antiderivative but the limits are 0 and a (or symmetric, or a and b), and replacing x by a − x (or a + b − x) produces a related expression, add the two forms.

Do I always need + C?

Yes, for every indefinite integral; omitting it can cost a mark. Definite integrals do not need it.

How many marks does integration carry?

It is part of Calculus (35 of 80 marks); expect integrals in Sections A, B, C and D.

Based on: CISCE ISC 2027 Mathematics syllabus; CISCE ISC 2027 Mathematics specimen question paper and answer key; CISCE Analysis of Pupil Performance, ISC Mathematics, 2025; CISCE Analysis of Pupil Performance, ISC Mathematics, 2024