Indefinite Integral Formulas

The main motto behind providing the Indefinite Integral Formulae here is to simplify your work while doing complex problems involving Integrals. You can use the simple formulas for Indefinite Integral and apply them in your calculations and get the solution easily. You will very well know the concepts by referring to the Antiderivative Formulas provided.

1. Integration of a function

∫f(x) dx = Φ(x) + c ⇔ \(\frac{d}{d x}\)[Φ(x)] = f(x)

2. Basic theorems on integration

If f(x), g(x) are two functions of a variable x and k is a constant, then

• ∫Kf(x)dx = K∫f(x)dx + c
• ∫[f (x) ± g(x)]dx = ∫f(x)dx ± ∫g(x)dx + c
• \(\frac{d}{d x}\)(∫f(x)dx)= f(x)
• ∫\(\left(\frac{d}{d x} f(x)\right)\)dx = f(x)

3. Standard Integrals

• ∫0.dx = c
• ∫1 .dx = x + c
• ∫kdx = kx + c, (k ∈ R)
• ∫xⁿdx = \(\frac{x^{n+1}}{n+1}\) + c, (n ≠ -1)
• ∫\(\frac{1}{x}\)dx = log_ex + c
• ∫e^xdx = e^x + c
• ∫a^xdx = \(\frac{\mathrm{a}^{\mathrm{x}}}{\log _{\mathrm{e}} \mathrm{a}}\) + c = a^xlog_ae + c
• ∫sin x dx = – cos x + c
• ∫cos x dx = sin x + c
• ∫tan x dx = log sec x + c = – log cos x + c
• ∫cotx dx= log sin x + c
• ∫secx dx = log (sec x + tan x) + c
  = – log (sec x – tan x) + c
  = log tan \(\left(\frac{\pi}{4}+\frac{x}{2}\right)\) + c

4. Integration by substitution

(i) When Integrand is a function of function
i.e. ∫ f[Φ(x)]Φ'(x)dx
Here we put Φ(x) = t so that Φ'(x) dx = dt and in that case the integrand is reduced to ∫ f (t)dt.

(ii) When integrand is the product of two factors such that one is the derivative of the other i.e.
1= ∫ f'(x)f(x)dx.
In this case we put f(x) = t and convert it into a standard integral.

(iii) Integral of a Function of the Form f (ax + b)
Here we put ax + b = t and convert it into standard integral. Obviously
if ∫ f(x)dx = Φ(x), then ∫ f(ax + b)dx = \(\frac{1}{a}\) Φ(ax+b) + c

(iv) Standard Form of Integrals

• \(\int \frac{f^{\prime}(x)}{f(x)}\)dx = log[f(x)] + c
• [f(x)]ⁿ f'(x)dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + c (provided n ≠ -1)
• \(\int \frac{f^{\prime}(x)}{\sqrt{f(x)}} d x=2 \sqrt{f(x)}+c\)

(v) Integral of the Form ∫\(\frac{d x}{a \sin x+b \cos x}\)
a sin x + b cos x putting a = r cos θ and b = r sin θ, we get
I = ∫\(\frac{d x}{r \sin (x+\theta)}\) = \(\frac{1}{r}\) ∫cosec(x + θ)dx
= \(\frac{1}{r}\) log tan\(\left(\frac{x}{2}+\frac{\theta}{2}\right)\) + c = \(\frac{1}{\sqrt{a^{2}+b^{2}}}\) log tan \(\left(\frac{x}{2}+\frac{1}{2} \tan ^{-1} \frac{b}{a}\right)\) + c

(vi) Standard Substitutions
Following standard substitutions will be useful-

Integral Form	Substitution
(a) \( \sqrt{a^{2}-x^{2}} \text { or } \frac{1}{\sqrt{a^{2}-x^{2}}}\)	x = a sin θ
(b) \( \sqrt{x^{2}+a^{2}} \text { or } \frac{1}{\sqrt{x^{2}+a^{2}}}\)	x = a tan θ or x = a sinh θ
(c) \( \sqrt{x^{2}-a^{2}} \text { or } \frac{1}{\sqrt{x^{2}-a^{2}}}\)	x = a sec θ or x = a cosh θ
(d) \( \sqrt{\frac{x}{a+x}} \text { or } \sqrt{\frac{a+x}{x}} \text { or } \sqrt{x(a+x)} \text { or } \frac{1}{\sqrt{x(a+x)}}\)	x = a tan2 θ
(e) \( \sqrt{\frac{x}{a-x}} \text { or } \sqrt{\frac{a-x}{x}} \text { or } \sqrt{x(a-x)} \text { or } \frac{1}{\sqrt{x(a-x)}}\)	x = a sin2 θ
(f) \( \sqrt{\frac{x}{x-a}} \text { or } \sqrt{\frac{x-a}{x}} \text { or } \sqrt{x(x-a)} \text { or } \frac{1}{\sqrt{x(x-a)}}\)	x = a sec2 θ
(g) \( \sqrt{\frac{a-x}{a+x}} \text { or } \sqrt{\frac{a+x}{a-x}}\)	x = a cos 2θ
(h) \( \sqrt{\frac{x-\alpha}{\beta-x}} \text { or } \sqrt{(x-\alpha)(\beta-x)}\), (β > α)	x = α cos2 θ + β sin2 θ

5. Integration by parts

(i) If u and v are two functions of x then
∫(u.v)dx = u(∫v dx) – ∫\(\left(\frac{d u}{d x}\right)\).(∫v dx)dx.
From the first letter of the words Inverse circular, Logarithmic, Algebraic, Trigonometric, Exponential functions, we get a word ILATE. Therefore first arrange the functions in the order according to letters of this word and then integrate by parts.

(ii) If the integral is of the form ∫e^x [f(x) + f'(x)] dx
∫e^x [f (x) + f'(x)]dx = e^x f(x) + c

(iii) If the integral is of the form ∫ [xf’ (x) + f (x)] dx
∫ [x f'(x) + f(x)]dx = x f (x) + c

6. Integration of Fractional Function

(i) When denominator can not be factorised:
(a) In this case integral may be in the form ∫\(\frac{d x}{a x^{2}+b x+c}\)
Here taking coefficient of x2 common from denominator, write
x² + (\(\frac{b}{a}\)) + \(\frac{c}{a}\) = (x + \(\left(x+\frac{b}{2 a}\right)^{2}-\frac{b^{2}-4 a c}{4 a^{2}}\)
Now the integrand so obtained can be evaluated easily by using standard formulas.

(b) \(\int \frac{(p x+q)}{a x^{2}+b x+c} d x\)
Here suppose that
px + q = A [diff. coefficient of
(ax² + bx + c)] + B = A (2ax + b) + B …(1)
Now comparing coefficient of x and constant terms,
we get A = p/2a, B = q – (pb/2a)
I = \(\frac{p}{2 a} \int \frac{2 a x+b}{a x^{2}+b x+c}\)dx + \(\left(q-\frac{p b}{2 a}\right) \int \frac{d x}{a x^{2}+b x+c}\)
Now we can integrate it easily.

(ii) Integration of rational functions containing only even powers of x. To find integral of such functions, first we divide numerator and denominator by x², then express numerator as
d\(\left(x \pm \frac{1}{x}\right)\) and denominator as a function of \(\left(x \mp \frac{1}{x}\right)\)

7. Integration of Irrational Functions

(i) If integral is in the form of \(\int \frac{d x}{\sqrt{a x^{2}+b x+c}}, \int \sqrt{a x^{2}+b x+c} d x\)
then we integrate it by expressing ax² + bx + c = (x + α)² + β

(ii) If the integrals of the form \(\int \frac{p x+q}{\sqrt{a x^{2}+b x+c}} d x\),
∫(px + q) \(\sqrt{a x^{2}+b x+c}\) dx. First we express px + q in the form px + q = A\(\left\{\frac{d}{d x}\left(a x^{2}+b x+c\right)\right\}\) + B and then proceed as usual with standard form.

8. Integration of Trigonometric Functions

Type – I

• ∫\(\frac{d x}{a+b \sin ^{2} x}\)
• ∫\(\frac{d x}{a+b \cos ^{2} x}\)
• ∫\(\frac{d x}{a \cos ^{2} x+b \sin x \cos x+c \sin ^{2} x}\)
• ∫\(\frac{d x}{(a \sin x+b \cos x)^{2}}\)

Divide numerator and denominator by cos² x in all such type of integrals and then put tanx = t.

Type – II

• ∫\(\frac{d x}{a+b \cos x}\)
• ∫\(\frac{d x}{a+b \sin x}\)
• ∫\(\frac{d x}{a \cos x+b \sin x}\)
• ∫\(\frac{d x}{a \sin x+b \cos x+c}\)

In such types of integrals we use following formulae for sin x and cos x in terms of tan (x/2).
sinx = \(\frac{2 \tan \left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)}\), cos x = \(\frac{1-\tan ^{2}\left(\frac{x}{2}\right)}{1+\tan ^{2}\left(\frac{x}{2}\right)}\) and then take tan (x/2) = t and integrate

Type – III

• ∫\(\frac{p \sin x+q \cos x}{a \sin x+b \cos x}\)
• ∫\(\frac{p \sin x}{a \sin x+b \cos x}\)
• ∫\(\frac{q \cos x}{a \sin x+b \cos x}\)

For their integration, we first express Nr. as follows-
Nr = A (Dr) + B (derivative of Dr)
Then integral = Ax + B log (Dr) + C

9. Integration of some important functions

Case -1
(a) \(\int \frac{x^{2} d x}{x^{4}+k x^{2}+a^{4}}=\frac{1}{2} \int \frac{\left(x^{2}+a^{2}\right) d x}{\left(x^{4}+k x^{2}+a^{4}\right)}+\frac{1}{2} \int \frac{x^{2}-a^{2}}{\left(x^{4}+k x^{2}+a^{4}\right)}\)
(b) \(\int \frac{d x}{\left(x^{4}+k x^{2}+a^{4}\right)}=\frac{1}{2 a^{2}} \int \frac{\left(x^{2}+a^{2}\right) d x}{\left(x^{4}+k x^{2}+a^{4}\right)}-\frac{1}{2 a^{2}} \int \frac{x^{2}-a^{2}}{\left(x^{4}+k x^{2}+a^{4}\right)}\)
for further integration use

• t = x – \(\frac{a^{2}}{x}\) or
• t = x + \(\frac{a^{2}}{x}\)

Case – II
(A) ∫\(\frac{d x}{(p x+q) \sqrt{a x+b}}\), put ax + b = t²
(B) ∫\(\frac{d x}{\left(p x^{2}+q x+c\right) \sqrt{a x+b}}\), put ax + b = t²
(C) ∫\(\frac{d x}{(p x+\theta)^{a} \sqrt{a x^{2}+b x+c}}\), put px + θ = \(\frac{1}{t}\) where α is + ve integer
(D) ∫\(\frac{d x}{\left(p x^{2}+q x+r\right) \sqrt{a x^{2}+b x+c}}\), put \(\frac{a x^{2}+b x+c}{p x^{2}+q x+r}\) = t²
(E) Integral of type ∫x^m (a + bxⁿ)^p dx where m, n, p ∈ Q

• If p ∈ N then expand (a + bxⁿ)^p then integrate
• If p < 0 then put x = t^α where α is common denominator of fractional part of m and n
• If \(\frac{m+1}{p}\) = Integer, then we put a + bxⁿ = t where α is the denominator of the fraction p
• If \(\frac{m+1}{n}\) + p = integer then substitution is a + bxⁿ = t^αxⁿ
  where α is the denominator of fraction “p”.

(F) ∫ sin^m x cosⁿ x dx to integrate

• If m and n are even integer then convert in terms of multiple angle
• m is odd and n is even then put cos x = t
• m is even, n is odd then put sin x = t
• If m, n both odd then put either sinx = t or cosx = t
• If m + n = – ve and even, then rearrange Q, in terms of “tanx” and put t = tanx
• ∫ sin^mx cos nx dx = \(\frac{\sin ^{m} x \sin n x}{(m+n)}-\frac{m}{(m+n)}\)
  ∫sin^m-1 x cos(n – 1)x dx
• ∫ cos^mx cos nx dx
  = \(\frac{\cos ^{m} x \sin ^{n} x}{m+n}\) + \(\frac{m}{(m+n)}\)∫ cos^m-1x cos (n – 1) dx
• ∫ sin^mx sin nx dx

= \(\frac{\sin ^{m} x+\cos n x}{(m+n)}+\frac{m}{(m+n)}\)∫ sin^m-1x sin (n – 1) dx
Imp. Note: If a function can be expressed in terms of elementary j function (formular format) then only it is integrable, other wise cannot For example:-
\(\int \mathrm{e}^{\sin x} \mathrm{dx}, \int \sqrt{\sin x} \mathrm{d} x, \int \frac{x^{4}}{x^{10}+1}, \int \frac{\cos x}{x} d x, \int \frac{d x}{\ell n \sin x} \text { etc. }\)

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