Indefinite Integration — MHT-CET Maths Formulas and Traps

Foundations — Antiderivatives, the +C, and Standard Formulae

Antiderivative and the Constant of Integration

Indefinite integral

\int f(x)\,dx = F(x) + C \quad\text{where}\quad F'(x) = f(x)

$F(x)$ any one antiderivative of $f$
$C$ arbitrary constant of integration

The Standard-Formula Table

Power rule (the most-used row)

\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C \quad (n \neq -1)

$n \neq -1$ the exclusion that makes $\int x^{-1} = \log|x|$ a separate row

The Linear-Argument Rule (replace x with ax + b)

Linear-argument rule

\int f(ax+b)\,dx = \dfrac{1}{a}\,F(ax+b) + C \quad\text{where}\quad \int f(x)\,dx = F(x)+C

$a$ coefficient of $x$ inside — you divide by it
$ax+b$ the argument; must be LINEAR for the shortcut to hold

Linearity and Algebraic Pre-processing

Linearity of integration

\int [a\,f(x) + b\,g(x)]\,dx = a\!\int f(x)\,dx + b\!\int g(x)\,dx

Finding C from a Boundary Condition

Solving for the constant

f(x) = F(x) + C,\qquad C = f(a) - F(a)

$f(a) = k$ the given boundary value
$F(a)$ antiderivative evaluated at the boundary point

Reconstruct the Function, Then Integrate

Composition of a function with itself

(f\circ f)(x) = f\big(f(x)\big)

Trigonometric Simplification Toolkit

The collapses you reach for most

1+\cos x = 2\cos^2\tfrac{x}{2},\quad 1-\cos x = 2\sin^2\tfrac{x}{2},\quad \sqrt{1\pm\sin 2x}=|\sin x\pm\cos x|

$\tfrac{x}{2}$ half-angle — appears whenever you collapse $1\pm\cos x$
$|\cdots|$ the root of a perfect square is a MODULUS; fix the sign on the given interval

Common traps

Never drop the +C on an indefinite integral

An indefinite integral with no

+C

is incomplete. MHT-CET options are written so that the 'no constant' version and a wrong-constant version both appear — only

+C

(or

+k

) is correct.

The power rule excludes $n = -1$

\int x^{-1}\,dx

is NOT

\dfrac{x^0}{0}

— that is undefined. It is the special row

\int \dfrac{1}{x}\,dx = \log|x| + C

. This exclusion is tested directly.

Only LINEAR insides get the 1/a shortcut

The rule holds because

\dfrac{d}{dx}(ax+b) = a

is a constant. For a non-linear inside like

x^2+1

or

\sqrt{x}

, the derivative is not constant — you must do a full substitution, not just divide by a number.

Divide before you integrate an improper fraction

If the numerator's degree is

\geq

the denominator's, you cannot jump to a log or arctan form. Polynomial-divide first; the remainder term is what carries the log/arctan.

Solve for C only AFTER integrating

The boundary value applies to

f

, not

f'

. Integrate fully first, keep the

C

, then substitute the known point. Substituting into

f'

does nothing.

Build f explicitly before integrating

You cannot integrate

f(g(x))

until you know

f

. Substitute to find the explicit rule first; integrating the composition blind is the most common error on these.

$1\pm\cos x$ (half-angle) vs $1\pm\cos 2x$ (power-reduction)

Different collapses:

1-\cos x = 2\sin^2\tfrac{x}{2}

but

1-\cos 2x = 2\sin^2 x

. Read the angle inside the cosine before choosing the factor — the wrong one halves or doubles the argument.

The root of a perfect square is a MODULUS

\sqrt{(\sin x-\cos x)^2} = |\sin x-\cos x|

, not

\sin x-\cos x

. Resolve the sign on the given interval: on

(\tfrac\pi4,\tfrac\pi2)

it is

+(\sin x-\cos x)

; on

(0,\tfrac\pi4)

it is

-(\sin x-\cos x)

. Dropping the modulus is the most common error on

\sqrt{1\pm\sin 2x}

problems.

$\sec\pm\tan$ — mind which way the half-angle shifts

\sec x+\tan x=\tan\!\left(\tfrac\pi4+\tfrac x2\right)

but

\sec x-\tan x=\tan\!\left(\tfrac\pi4-\tfrac x2\right)

. Useful check:

(\sec x+\tan x)(\sec x-\tan x)=\sec^2x-\tan^2x=1

.

Integration by Substitution — the Workhorse Method

Learn this subtopic in the notes

The Substitution Rule

Substitution rule

\int f(g(x))\,g'(x)\,dx = \int f(u)\,du,\quad u = g(x)

$u = g(x)$ the inner function you rename
$du = g'(x)\,dx$ its differential, which must appear in the integrand

The f'(x)/f(x) → log Pattern

Logarithmic integral

\int \dfrac{f'(x)}{f(x)}\,dx = \log|f(x)| + C

$f(x)$ the denominator
$f'(x)$ its derivative — must equal the numerator (up to a constant)

Power of a Function times its Derivative

Power-of-a-function rule

\int [f(x)]^n\,f'(x)\,dx = \dfrac{[f(x)]^{n+1}}{n+1} + C \quad (n \neq -1)

Root and Linear-Radical Substitutions

Linear-radical substitution

t = \sqrt{ax+b}\ \Rightarrow\ x = \dfrac{t^2 - b}{a},\quad dx = \dfrac{2t}{a}\,dt

Reciprocal and Take-out-the-Power Substitutions

The reciprocal substitution

t = x \pm \dfrac{k}{x} \;\Rightarrow\; dt = \left(1 \mp \dfrac{k}{x^2}\right)dx

$\pm k/x$ sign chosen so $dt$ matches the numerator after dividing by $x^2$

Exponential and Special Substitutions

Exponential substitution

t = e^{x}\ \Rightarrow\ dt = e^{x}\,dx

Common traps

Adjust for the missing constant

If

du = g'(x)\,dx

appears up to a numeric factor (e.g. you have

x\,dx

but

du = 2x\,dx

), pull the constant out:

x\,dx = \tfrac12\,du

. Forgetting the

\tfrac12

is the most common slip.

Engineer the numerator into f'(x) + leftover

Rarely is the top exactly

f'(x)

. Write it as 'constant

\times f'(x)

+ remainder', send the first piece to a clean log, and handle the remainder separately. The whole-number coefficient comes from matching.

n = −1 is NOT this rule

If the power is

-1

(i.e.

\int f'/f

), the power rule blows up. That case is the log pattern. Every other integer/fraction power uses

\dfrac{u^{n+1}}{n+1}

.

Re-express EVERY x, including dx

After

t = \sqrt{x}

, both the integrand AND

dx = 2t\,dt

must be rewritten. Leaving a stray

x

or the old

dx

behind is the classic substitution error.

Pick the sign of t = x ± k/x from the numerator

After dividing by

x^2

, if the numerator is

1 - \dfrac{k}{x^2}

use

t = x + \dfrac{k}{x}

; if it is

1 + \dfrac{k}{x^2}

use

t = x - \dfrac{k}{x}

. The substitution only works when

dt

reproduces the numerator exactly.

Towers: substitute the INNER exponential

For

\int a^{a^x} a^x\,dx

, the right substitution is

u = a^x

, not

u = a^{a^x}

. The stray

a^x\,dx

is what becomes

du

(up to

\log a

).

Trigonometric Integrals I — Powers and Identities

Learn this subtopic in the notes

The Standard tan, cot, sec, cosec Integrals

The two that need the conjugate trick

\int \sec x\,dx = \log|\sec x + \tan x| + C,\qquad \int \csc x\,dx = \log|\csc x - \cot x| + C

Power Reduction with Pythagorean Identities

Key reduction identity

\tan^2 x = \sec^2 x - 1

Identity Simplification before Integrating

A workhorse collapse

\tan x + \cot x = \dfrac{\sin^2 x + \cos^2 x}{\sin x\cos x} = \dfrac{2}{\sin 2x} = 2\csc 2x

Simplify the Inverse-Trig Argument First

Cancel, then integrate

\tan^{-1}(\tan\theta) = \theta,\quad \sin^{-1}(\sin\theta)=\theta \quad (\theta\text{ in principal range})

Common traps

∫sec and ∫cosec are NOT plain logs of sec/cosec

\int \sec x\,dx = \log|\sec x + \tan x|

, not

\log|\sec x|

. The

+\tan x

(and

-\cot x

for cosec) is exactly what the conjugate trick produces — options drop it to bait you.

Keep one sec²x to pair with the tan-power

The whole method works because

\sec^2 x\,dx = d(\tan x)

. Each reduction must leave a

\sec^2 x

attached to a power of

\tan x

so the power rule applies — otherwise you stall.

Try an identity before a substitution

Reaching for Weierstrass on

\int(\tan x + \cot x)\,dx

is a long detour — one identity makes it

2\csc 2x

instantly. Simplify first; substitute only if no identity collapses it.

Reduce the argument BEFORE integrating

Never reach for by-parts on

\int\tan^{-1}(\cdots)\,dx

until you've tried to collapse the inside. If the argument is a half/double-angle form, the inverse cancels and the integral is a one-line polynomial — by-parts is a needless detour.

Rational Functions and Partial Fractions

Learn this subtopic in the notes

Standard Quadratic Denominator Forms

The arctan form

\int \dfrac{dx}{x^2 + a^2} = \dfrac{1}{a}\tan^{-1}\dfrac{x}{a} + C

Completing the square

ax^2 + bx + c = a\left(x + \dfrac{b}{2a}\right)^2 + \left(c - \dfrac{b^2}{4a}\right)

Linear Numerator over a Quadratic (Numerator Split)

Numerator as derivative-of-denominator plus constant

px + q = A\dfrac{d}{dx}(ax^2+bx+c) + B

$A$ coefficient that reproduces the $x$ -term via $(ax^2+bx+c)'$
$B$ leftover constant — its integral completes the square

Partial-Fraction Decomposition

Distinct-linear decomposition

\dfrac{px + q}{(x-a)(x-b)} = \dfrac{A}{x-a} + \dfrac{B}{x-b}

Common traps

x⁴ + bx² + c → try t = x ± k/x

A denominator quadratic in

x^2

with a matching numerator

(1 \mp k/x^2)

is the signal for

t = x \pm k/x

. It collapses the quartic to a simple

t^2 + 1

arctan — a recurring MHT-CET shape.

Factor the sign on x² before completing the square

With

-x^2

terms, pull out the

-1

first:

7 - 6x - x^2 = -(x^2 + 6x - 7)

. Forgetting the sign flip turns an arcsin into a (wrong) log or vice versa.

Split the numerator BEFORE completing the square

The derivative-piece must come out first (it gives the

\sqrt{}

or log term). Only the leftover constant goes through completing the square. Completing the square first leaves the

x

in the numerator with nowhere to go.

Improper fraction? Divide before decomposing

Partial fractions require the numerator degree to be LESS than the denominator's. If not, polynomial-divide first, then decompose the proper remainder.

Repeated factor needs every power

For

(x-a)^2

you must include BOTH

\dfrac{A}{x-a}

and

\dfrac{B}{(x-a)^2}

. Dropping the first-power term gives an unsolvable system.

Trigonometric Integrals II — Rational Forms and Substitutions

Learn this subtopic in the notes

The Half-Angle (Weierstrass) Substitution

Weierstrass substitution

t = \tan\dfrac{x}{2}:\quad \sin x = \dfrac{2t}{1+t^2},\ \cos x = \dfrac{1-t^2}{1+t^2},\ dx = \dfrac{2\,dt}{1+t^2}

Divide by cos²x for a + b·sin²x Forms

After dividing by cos²x

\int \dfrac{\sec^2 x\,dx}{A + B\tan^2 x} \xrightarrow{\,t=\tan x\,} \int \dfrac{dt}{A + Bt^2}

Product of Two Shifted Sines (or Cosines)

Shifted-angle split

\dfrac{1}{\sin(x-a)\sin(x-b)} = \dfrac{1}{\sin(b-a)}\big[\cot(x-b) - \cot(x-a)\big]

$\sin(b-a)$ constant divisor — the sine of the difference of the two shifts

Trig to Partial Fractions (substitute, then decompose)

The bridge

\int R(\sin x)\,\cos x\,dx \;\xrightarrow{\,t=\sin x\,}\; \int R(t)\,dt \;\;(\text{now rational — decompose})

The Fractional-Power tan Trick

The reduction (m + n even)

\int \sin^m x\,\cos^n x\,dx \xrightarrow{\,t=\tan x\,} \int t^{m}\,(1+t^2)^{\frac{m+n}{2}-1}\,dt

Numerator as Denominator + its Derivative

Decomposition of the numerator

\text{num} = A\cdot\text{den} + B\cdot(\text{den})' \;\Rightarrow\; \int\dfrac{\text{num}}{\text{den}}\,dx = Ax + B\log|\text{den}| + C

Common traps

Weierstrass is for a + b·sin/cos, not a + b·sin²

If the denominator has

\sin^2 x

or

\cos^2 x

(an even power), the half-angle substitution gives a messy quartic. Use divide-by-

\cos^2 x

instead — the next concept.

Three sin/cos denominators, three substitutions

Denominator has

\sin^2/\cos^2

(even, plain angle): divide by

\cos^2 x

, use

t=\tan x

. Denominator has

\sin 2x/\cos 2x

(double angle): use

t=\tan x

with the

2x

formulae. Denominator has plain

\sin x/\cos x

(odd, single power): use Weierstrass

t=\tan(x/2)

. Picking the wrong one makes the algebra explode.

Split it — don't reach for Weierstrass

Weierstrass on a product of two shifted sines produces a quartic in

t

and stalls. The shifted-angle split is one identity and one line. Keep the constant

\sin(b-a)

out front — it is NOT a function of

x

.

No spare cos/sin → no bridge

The substitution only works if exactly one

\cos x\,dx

(or

\sin x\,dx

) is available to become

dt

and everything else turns rational in

t

. If both

\sin x

and

\cos x

appear in odd/mixed ways with nothing spare, fall back to Weierstrass.

Check m + n is an even integer first

The trick only collapses cleanly when

m + n

is an even integer (so

\cos^{m+n}x

becomes an integer power of

\sec^2 x

). If it is odd, this route leaves a stray

\sec x

or

\cos x

and you need a different method.

Convert tan-fractions to sin/cos first

A ratio in

\tan x

is easiest after multiplying through by

\cos x

to get a sin/cos ratio — then the 'denominator + its derivative' decomposition is clean.

Integration by Parts

Learn this subtopic in the notes

Integration by Parts and the LIATE Rule

Integration by parts

\int u\,dv = uv - \int v\,du

$u$ factor to differentiate (earliest in LIATE)
$dv$ factor to integrate (the rest, including $dx$ )

Cyclic Integrals (Return-to-Self)

The cyclic result

\int e^{ax}\sin bx\,dx = \dfrac{e^{ax}(a\sin bx - b\cos bx)}{a^2 + b^2} + C

The eˣ[f(x) + f'(x)] Family

The eˣ[f + f'] shortcut

\int e^x\big[f(x) + f'(x)\big]\,dx = e^x f(x) + C

$f(x)$ the function whose value lands in the answer
$f'(x)$ its derivative — must be the other half of the bracket

Integrals of √(quadratic) — Standard Results (syllabus reference)

Square root of a quadratic — the three results

\int \sqrt{a^2 - x^2}\,dx = \dfrac{x}{2}\sqrt{a^2 - x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a} + C

$a^2 - x^2$ arcsin result
$a^2 + x^2,\ x^2 - a^2$ log results (signs differ)

Generalised (Tabular) By-Parts — a shortcut

Tabular by-parts series

\int u\,v\,dx = u\,v_1 - u'\,v_2 + u''\,v_3 - u'''\,v_4 + \cdots

$u', u'', \ldots$ successive derivatives of the polynomial $u$
$v_1, v_2, \ldots$ successive integrals of $v$

Common traps

A lone log or inverse-trig still uses parts

\int \log x\,dx

and

\int \tan^{-1}x\,dx

look like single functions, but they are integrated by parts with

dv = dx,\ v = x

. There is no direct formula.

Stop after two rounds — don't loop forever

The point of a cyclic integral is that the original

I

reappears after two rounds. Recognise it and solve algebraically; a third by-parts just sends you in circles.

Use identities to expose f + f'

The bracket rarely arrives as a clean

f + f'

. Apply identities first — e.g.

1 + \cot^2 = \csc^2

— so that one term is a function and the other is its exact derivative. Then the answer is immediate.

Syllabus result, not a current bank pattern

This bank's recent CET questions keep the square root in a DENOMINATOR (handled by the numerator-split + completing-the-square concepts). The

\int\sqrt{\text{quadratic}}\,dx

form here is on the syllabus and could appear — but don't expect it among the tagged drills, because the live bank has none yet.

Only for a polynomial first function

The shortcut relies on the polynomial's derivatives reaching zero. If neither factor is a polynomial (e.g.

\int e^x\sin x\,dx

), the table never terminates — use ordinary or cyclic by-parts instead.

Indefinite Integration formulas

Foundations — Antiderivatives, the +C, and Standard Formulae

Antiderivative and the Constant of Integration

The Standard-Formula Table

The Linear-Argument Rule (replace x with ax + b)

Linearity and Algebraic Pre-processing

Finding C from a Boundary Condition

Reconstruct the Function, Then Integrate

Trigonometric Simplification Toolkit

Common traps

Integration by Substitution — the Workhorse Method

The Substitution Rule

The f'(x)/f(x) → log Pattern

Power of a Function times its Derivative

Root and Linear-Radical Substitutions

Reciprocal and Take-out-the-Power Substitutions

Exponential and Special Substitutions

Common traps

Trigonometric Integrals I — Powers and Identities

The Standard tan, cot, sec, cosec Integrals

Power Reduction with Pythagorean Identities

Identity Simplification before Integrating

Simplify the Inverse-Trig Argument First

Common traps

Rational Functions and Partial Fractions

Standard Quadratic Denominator Forms

Linear Numerator over a Quadratic (Numerator Split)

Partial-Fraction Decomposition

Common traps

Trigonometric Integrals II — Rational Forms and Substitutions

The Half-Angle (Weierstrass) Substitution

Divide by cos²x for a + b·sin²x Forms

Product of Two Shifted Sines (or Cosines)

Trig to Partial Fractions (substitute, then decompose)

The Fractional-Power tan Trick

Numerator as Denominator + its Derivative

Common traps

Integration by Parts

Integration by Parts and the LIATE Rule

Cyclic Integrals (Return-to-Self)

The eˣ[f(x) + f'(x)] Family

Integrals of √(quadratic) — Standard Results (syllabus reference)

Generalised (Tabular) By-Parts — a shortcut

Common traps

More MHT-CET Maths formula sheets