Important Integration Formulas for JEE Maths - Indefinite and Definite Formulas List

In this section, students will learn the main indefinite and definite integration formulas as well as some main properties of integration. In general, integration is the reverse operation of differentiation. It is also called antiderivative. The formulas provided here will help students to easily remember them for the exam and score higher marks in the exams.

Integration Formulas

Indefinite Integration:

If f and g are functions of x such that g’(x) = f(x) then,

\begin{array}{l} \int f (x) d x = g (x) + c \Leftrightarrow \frac{d}{d x} {g (x) + c} = f (x) \end{array}

Here, c is called the constant of integration.

Standard formula:

\begin{array}{l} (a) \int x^{n} d x = \frac{x^{n + 1}}{n + 1} + c, n \neq - 1 \end{array}

\begin{array}{l} (b) \int \frac{1}{x} d x = \log_{e} | x | + c \end{array}

(c) ∫e^x dx = e^x+c

\begin{array}{l} (d) \int a^{x} d x = \frac{a^{x}}{\log_{e} a} + c \end{array}

(e) ∫ sin x dx = -cos x + c

(f) ∫ cos x dx = sin x + c

(g) ∫ sec² x dx = tan x + c

(h) ∫ cosec² x dx = -cot x + c

(h) ∫ sec x tan x dx = sec x + c

(i) ∫ cosec x cot x dx = – cosec x + c

(j) ∫ cot x dx = log|sin x| + c

(k) ∫ tan x dx = -log|cos x| + c

(l) ∫ sec x dx = log |sec x + tan x| + c

(m) ∫ cosec x dx = log |cosec x – cot x| + c

\begin{array}{l} (n) \int \frac{1}{\sqrt{a^{2} - x^{2}}} d x = \sin^{- 1} (\frac{x}{a}) + c \end{array}

\begin{array}{l} (o) \int - \frac{1}{\sqrt{a^{2} - x^{2}}} d x = \cos^{- 1} (\frac{x}{a}) + c \end{array}

\begin{array}{l} (p) \int \frac{1}{a^{2} + x^{2}} d x = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) + c \end{array}

\begin{array}{l} (q) \int - \frac{1}{a^{2} + x^{2}} d x = \frac{1}{a} \cot^{- 1} (\frac{x}{a}) + c \end{array}

\begin{array}{l} (r) \int \frac{1}{x \sqrt{x^{2} - a^{2}}} d x = \frac{1}{a} \sec^{- 1} (\frac{x}{a}) + c \end{array}

\begin{array}{l} (s) \int - \frac{1}{x \sqrt{x^{2} - a^{2}}} d x = \frac{1}{a} c o s e c^{- 1} (\frac{x}{a}) + c \end{array}

If in place of x we have (ax+b), then the same formula is applicable but we must divide by coefficient of x or derivative of (ax+b), i.e., a.

\begin{array}{l} (i) \int (a x + b)^{n} d x = \frac{(a x + b)^{n + 1}}{a (n + 1)} + c, n \neq - 1 \end{array}

\begin{array}{l} (i i) \int \frac{d x}{a x + b} = \frac{1}{a} l n (a x + b) + c \end{array}

(iii) ∫ e^ax+b dx = (1/a) e^ax+b+c

\begin{array}{l} (i v) \int a^{p x + q} d x = \frac{1}{p} \frac{a^{p x + q}}{\ln a} + c, a > 0 \end{array}

(v) ∫ sin(ax + b)dx = -(1/a) cos(ax + b) + c

(vi) ∫ cos(ax + b) dx = (1/a) sin(ax + b) + c

(vii) ∫ tan(ax + b)dx = (1/a) ln sec(ax + b) + c

(viii) ∫ cot(ax+b)dx = (1/a) ln sin (ax + b) + c

(ix) ∫ sec²(ax+b)dx = (1/a) tan(ax+b) + c

(x) ∫ cosec² (ax+b) dx = -(1/a) cot (ax+b) + c

(xi) ∫ sec (ax+b) ⋅ tan (ax+b) dx = (1/a) sec (ax+b) + c

(xii) ∫ cosec (ax+b) ⋅ cot (ax+b) dx = -(1/a) cosec (ax+b) + c

(xiii) ∫ sec x dx = ln (sec x + tan x) + c or

\begin{array}{l} = \ln t a n (\frac{π}{4} + \frac{x}{2}) \end{array}

(xiv) ∫ cosec x dx = ln (cosec x – cot x) + c or ln tan (x/2) +c or –ln (cosec x + cot x)+ c

\begin{array}{l} (x v) \int \frac{d x}{\sqrt{a^{2} - x^{2}}} = \sin^{- 1} \frac{x}{a} + c \end{array}

\begin{array}{l} (x v i) \int \frac{d x}{a^{2} + x^{2}} = \frac{1}{2} \tan^{- 1} \frac{x}{a} + c \end{array}

\begin{array}{l} (x v i i) \int \frac{d x}{| x | \sqrt{x^{2} - a^{2}}} = \frac{1}{a} \sec^{- 1} \frac{x}{a} \end{array}

\begin{array}{l} (x v i i i) \int \frac{d x}{\sqrt{x^{2} + a^{2}}} = \ln [x + \sqrt{x^{2} + a^{2}}] + c \end{array}

\begin{array}{l} (x i x) \int \frac{d x}{\sqrt{x^{2} - a^{2}}} = \ln [x + \sqrt{x^{2} - a^{2}}] + c \end{array}

\begin{array}{l} (x x) \int \frac{d x}{a^{2} - x^{2}} = \frac{1}{2 a} \ln | \frac{a + x}{a - x} | + c \end{array}

\begin{array}{l} (x x i) \int \frac{d x}{x^{2} - a^{2}} = \frac{1}{2 a} \ln | \frac{x - a}{x + a} | + c \end{array}

\begin{array}{l} (x x i i) \int \sqrt{a^{2} - x^{2}} d x = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} \sin^{- 1} \frac{x}{a} + c \end{array}

\begin{array}{l} (x x i i i) \int \sqrt{x^{2} + a^{2}} d x = \frac{x}{2} \sqrt{x^{2} + a^{2}} + \frac{a^{2}}{2} \ln (\frac{x + \sqrt{x^{2} + a^{2}}}{a}) + c \end{array}

\begin{array}{l} (x x i v) \int \sqrt{x^{2} - a^{2}} d x = \frac{x}{2} \sqrt{x^{2} - a^{2}} - \frac{a^{2}}{2} \ln (\frac{x + \sqrt{x^{2} - a^{2}}}{a}) + c \end{array}

\begin{array}{l} (x x v) \int e^{a x} . \sin b x d x = \frac{e^{a x}}{a^{2} + b^{2}} (a \sin b x - b \cos b x) + c \end{array}

\begin{array}{l} (x x v i) \int e^{a x} . \cos b x d x = \frac{e^{a x}}{a^{2} + b^{2}} (a \cos b x + b \sin b x) + c \end{array}

Theorems on integration:

(i) ∫ c f(x) dx = c ∫ f(x) dx

(ii) ∫[f(x) ± g(x)] dx = ∫f(x) dx ± ∫g(x) dx

(iii) ∫ f(x) dx = g(x)+c

\begin{array}{l} \Rightarrow \int f (a x + b) d x = \frac{g (a x + b)}{a} + c \end{array}

Integration by substitutions:

If we substitute f(x) = t, then f’(x) dx = dt

Integration by part:

\begin{array}{l} \int [f (x) g (x)] d x = f (x) \int g (x) d x - \int (\frac{d}{d x} (f (x)) \int (g (x)) d x) d x \end{array}

(i) when you find integral ∫g(x) dx then it will not contain an arbitrary constant.

(ii) ∫g(x) dx should be taken as the same in both terms.

(iii) the choice of f(x) and g(x) is decided by ILATE rule.

Integration of type

\begin{array}{l} \int \frac{d x}{a x^{2} + b x + c}, \int \frac{d x}{\sqrt{a x^{2} + b x + c}}, \int \sqrt{a x^{2} + b x + c} d x \end{array}

Make the substitution

\begin{array}{l} x + \frac{b}{2 a} = t \end{array}

Integration of type

(1)

\begin{array}{l} \int \frac{p x + q}{a x^{2} + b x + c} d x, \int \frac{p x + q}{\sqrt{a x^{2} + b x + c}} d x, \int (p x + q) \sqrt{a x^{2} + b x + c} d x \end{array}

Make the substitution

\begin{array}{l} x + \frac{b}{2 a} = t, \end{array}

then split the integral as the sum of two integrals, one containing the linear term and the other containing the constant term.

8. Integration of trigonometric functions

(1)

\begin{array}{l} \int \frac{d x}{a + b \sin^{2} x} or \int \frac{d x}{a + b \cos^{2} x} or \int \frac{d x}{a \sin^{2} x + b \sin x \cos x + c \cos^{2} x} \end{array}

put tan x = t.

(2)

\begin{array}{l} \int \frac{d x}{a + b \sin x} or \int \frac{d x}{a + b \cos x} or \int \frac{d x}{a + b \sin x + c \cos x} \end{array}

put tan (x/2) = t.

(3)

\begin{array}{l} \int \frac{a \cos x + b \sin x + c}{l \cos x + m \sin x + n} d x \end{array}

Express

\begin{array}{l} N r = A (D r) + B \frac{d}{d x} (D r) + c \end{array}

and proceed.

Integration of type ∫sin mx. cos nx dx

Case 1. If m and n are even natural numbers then express sin mx cos nx in the terms of sines and cosines of multiples of x by using trigonometric results or De’ Moivere’s theorem.

Case 2. If m is an odd natural number then put cos x = t.

If n is an odd natural number then put sin x = t.

If both m and n are odd natural numbers then put either sin x = t or cos x = t.

Case 3. When m+n is a negative even integer then put tan x = t.

Integration of type
$\begin{array}{l} \int \frac{x^{2} \pm 1}{x^{4} + K x^{2} + 1} d x \end{array}$
where K is any constant.

Divide numerator and denominator by x² and put

\begin{array}{l} x \pm \frac{1}{x} = t \end{array}

Integration of type
$\begin{array}{l} \int \frac{d x}{(a x + b) \sqrt{p x + q}} or \int \frac{d x}{(a x^{2} + b x + c) \sqrt{p x + q}} \end{array}$
Put px+q = t²

Integration of type

(i)

\begin{array}{l} \int \frac{d x}{(a x + b) \sqrt{p x^{2} + q x + r}} \end{array}

: put ax+b = 1/t

(ii)

\begin{array}{l} \int \frac{d x}{(a x^{2} + b) \sqrt{p x^{2} + q}} \end{array}

: put x = 1/t

Integration of type

(i)

\begin{array}{l} \int \sqrt{\frac{x - α}{β - x}} d x or \int \sqrt{(x - α) (β - x)} \end{array}

put x = α cos²θ + β sin²θ

(ii)

\begin{array}{l} \int \sqrt{\frac{x - α}{x - β}} d x or \int \sqrt{(x - α) (x - β)} \end{array}

put x = α sec²θ – β tan²θ

(iii)

\begin{array}{l} \int \frac{d x}{\sqrt{(x - α) (x - β)}} \end{array}

: put x – α = t² or x – β = t²

Reduction formula of ∫ tanⁿ x dx, ∫ cotⁿ x dx, ∫ secⁿ x dx, ∫ cosecⁿ x dx:

(i) If In = ∫ tanⁿ x dx, then

\begin{array}{l} l n = \frac{t a n^{n - 1} x}{n - 1} - I_{n - 2} \end{array}

(ii) If In = ∫ cotⁿx dx, then

\begin{array}{l} l n = - \frac{\cot^{n - 1} x}{n - 1} - I_{n - 2} \end{array}

(iii) If In = ∫ secⁿ x dx, then

\begin{array}{l} l n = \frac{t a n x \sec^{n - 2} x}{n - 1} + \frac{n - 2}{n - 1} I_{n - 2} \end{array}

(iv) If In = ∫ cosecⁿ x, then

\begin{array}{l} l n = \frac{\cot x c o s e c^{n - 2} x}{n - 1} + \frac{n - 2}{n - 1} I_{n - 2} \end{array}

Definite Integration:

Properties of definite integral:

\begin{array}{l} (1) \int_{a}^{b} f (x) d x = \int_{a}^{b} f (t) d t \end{array}

\begin{array}{l} (2) \int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x \end{array}

\begin{array}{l} (3) \int_{a}^{b} f (x) d x = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x \end{array}

\begin{array}{l} (4) \int_{- a}^{a} f (x) d x = \int_{0}^{a} (f (x) + f (- x)) d x \end{array}

\begin{array}{l} = {\begin{array}{c} 2 \int_{0}^{a} f (x) d x, & f (- x) = f (x) \\ 0, & f (- x) = - f (x) \end{array} \end{array}

\begin{array}{l} (5) \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x \end{array}

\begin{array}{l} (6) \int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x \end{array}

\begin{array}{l} (7) \int_{0}^{2 a} f (x) d x = \int_{0}^{a} (f (x) + f (2 a - x)) d x \end{array}

\begin{array}{l} = {\begin{array}{c} 2 \int_{0}^{a} f (x) d x, & f (2 a - x) = f (x) \\ 0, & f (2 a - x) = - f (x) \end{array} \end{array}

(8) If f(x) is a periodic function with period T, then

(i)

\begin{array}{l} \int_{0}^{n T} f (x) d x = n \int_{0}^{T} f (x) d x, n \in Z \end{array}

(ii)

\begin{array}{l} \int_{0}^{a + n T} f (x) d x = n \int_{0}^{T} f (x) d x, n \in Z, a \in R \end{array}

(iii)

\begin{array}{l} \int_{m t}^{n T} f (x) d x = (n - m) \int_{0}^{T} f (x) d x, m, n \in Z \end{array}

(iv)

\begin{array}{l} \int_{n t}^{a + n T} f (x) d x = \int_{0}^{a} f (x) d x, n \in Z, a \in R \end{array}

(9) If

\begin{array}{l} ψ (x) \leq f (x) \leq ϕ (x) f o r a \leq x \leq b t h e n \int_{a}^{b} ψ (x) d x \leq \int_{a}^{b} f (x) d x \leq \int_{a}^{b} ϕ (x) \end{array}

(10) If m≤ f(x) ≤ M for a≤x≤b, then

\begin{array}{l} m (b - a) \leq \int_{a}^{b} f (x) d x \leq M (b - a) \end{array}

\begin{array}{l} (11) | \int_{a}^{b} f (x) d x | \leq \int_{a}^{b} | f (x) | d x \end{array}

(12) If f(x) ≥0 on [a,b] then

\begin{array}{l} \int_{a}^{b} f (x) d x \geq 0 \end{array}

Leibnitz Theorem:

\begin{array}{l} If F (x) = \int_{g (x)}^{h (x)} f (t) d t, then \end{array}

\begin{array}{l} \frac{d F (x)}{d x} = h^{'} (x) f (h (x)) - g^{'} (x) f (g (x)) \end{array}

Definite integrals as a limit of sum:

\begin{array}{l} \int_{a}^{b} f (x) d x = lim_{n \to \infty} \sum_{r = 0}^{n - 1} h f (a + r h) = lim_{n \to \infty} \sum_{r = 0}^{n - 1} (\frac{b - a}{n}) f (a + \frac{(b - a) r}{n}) \end{array}

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