Indefinite Integral : Basic Concepts & Formulae

Basic Concepts :

Let F(x) be a differentiable function of x such that

$ \displaystyle \frac{d}{dx}[F(x)] = f(x) $

Then F(x) is called the integral of f(x).

Symbolically, it is written as $ \displaystyle \int f(x) dx = F(x) $

f(x) , the function to be integrated , is called the integrand.

F(x) is also called the anti-derivative (or primitive function) of f(x).

Constant of Integration:

As the differential coefficient of a constant is zero, we have

$ \displaystyle \frac{d}{dx}[F(x)] = f(x) $

$ \displaystyle \frac{d}{dx}[F(x) +c ] = f(x) $

Therefore, $ \displaystyle \int f(x) dx = F(x) + c $

This constant c is called the constant of integration and can take any real value

Properties of Indefinite Integration:

(i) $ \displaystyle \int a f(x) dx = a \int f(x) dx $ ( Here ‘ a ‘ is a constant)

(ii) $ \displaystyle \int ( f(x) + g(x) ) dx = \int f(x) dx + \int g(x) dx $

(iii) If $ \displaystyle \int f(u) du = F(u) + c $

Then $ \displaystyle \int f(ax+b) dx = \frac{1}{a}F(ax+b) +c \; , a\ne 0 $

Integration as the Inverse Process of Differentiation :

Evaluate :

(i) $ \displaystyle \int tan^2 x dx $

(ii) $ \displaystyle \int e^x (cosx-sinx) dx $

(iii) $ \displaystyle \int 3 x^{1/2} (1 + x^{3/2})dx $

(iv) $ \displaystyle \int sin3x cos2x dx $

Solutions: (i) $ \displaystyle \int tan^2 x dx $

$ \displaystyle = \int (sec^2 x -1)dx $

= tanx – x + c

(ii) $ \displaystyle \int e^x (cosx-sinx) dx $

Here ex(cosx − sinx) is the derivative of ex cosx.

=> I = excosx + c.

(iii) $ \displaystyle \int 3 x^{1/2} (1 + x^{3/2})dx $

Here 3x1/2(1 + x3/2) is the derivative of (1 + x3/2)2

=> I = (1 + x3/2)2 + c.

(iv) $ \displaystyle \int sin3x cos2x dx $

NOTE :  When solving such problems it is expedient to use the following trigonometric identities :

$ \displaystyle sinmx \; cos nx = \frac{1}{2}[sin(m-n)x + sin(m+n)x]$

$ \displaystyle sinmx \; sin nx = \frac{1}{2}[cos(m-n)x – cos(m+n)x]$

$ \displaystyle cos mx \; cos nx = \frac{1}{2}[cos(m-n)x + cos(m+n)x]$

Here $ \displaystyle sin3x \; cos 2x = \frac{1}{2}[sin5x + sin x]$

=> $ \displaystyle I = \frac{1}{2} \int (sin5x + sin x)dx $

$ \displaystyle = -\frac{1}{2} [\frac{cos5x}{5}+ cosx] + c $

Exercise :

Integrate the following functions

(i) $ \displaystyle \int \frac{(1+x)^3}{x} dx $

(ii) $ \displaystyle \int \frac{e^{3x}+e^{5x}}{e^x + e^{-x}} dx $

(iii) $ \displaystyle \int \frac{1 + tan^2 x}{1 + cot^2 x} dx $

(iv) $ \displaystyle \int cos^3 x dx $

Basic formulae :

$ \displaystyle \frac{d}{dx}(\frac{x^{n+1}}{n+1}) = x^n \Rightarrow \int x^n dx = \frac{x^{n+1}}{n+1}+c $

$ \displaystyle \frac{d}{dx}(ln|x|) = \frac{1}{x} \Rightarrow \int \frac{1}{x} dx = ln|x| + c $

$ \displaystyle \frac{d}{dx}(e^x) = e^x \Rightarrow \int e^x dx = e^x + c $

$ \displaystyle \frac{d}{dx}(a^x) = (a^x ln a ) \Rightarrow \int a^x dx = \frac{a^x}{ln a} + c $

$ \displaystyle \frac{d}{dx}(sinx) = cosx \Rightarrow \int cosx dx = sinx + c $

$ \displaystyle \frac{d}{dx}(cosx) = -sinx \Rightarrow \int sinx dx = -cosx + c $

$ \displaystyle \frac{d}{dx}(tanx) = sec^2x \Rightarrow \int sec^2 x dx = tanx + c $

$ \displaystyle \frac{d}{dx}(secx) = secx tanx \Rightarrow \int secx tanx dx = secx + c $

$ \displaystyle \frac{d}{dx}(cotx) = -cosec^2x \Rightarrow \int cosec^2 x dx = -cotx + c $

$ \displaystyle \frac{d}{dx}(sin^{-1}(\frac{x}{a}) = \frac{1}{\sqrt{a^2 -x^2}} \Rightarrow \int \frac{1}{\sqrt{a^2 -x^2}} dx = sin^{-1}(\frac{x}{a} + c $

$ \displaystyle \frac{d}{dx}(tan^{-1}(\frac{x}{a}) = \frac{a}{x^2 +a^2 } \Rightarrow \int \frac{1}{x^2 +a^2 } dx = \frac{1}{a}tan^{-1}(\frac{x}{a} + c $

$ \displaystyle \frac{d}{dx}(sec^{-1}x) = \frac{1}{|x|\sqrt{x^2 -1}} \Rightarrow \int \frac{1}{|x|\sqrt{x^2 -1}} dx = sec^{-1}x + c $

$ \displaystyle \int cotx dx = \int \frac{cosx}{sinx} = ln |sinx| + c $

$ \displaystyle \int tanx dx = \int \frac{sinx}{cosxx} = ln |secx| + c = -ln |cosx| + c $

$ \displaystyle \int secx dx = \int \frac{secx (secx+tanx)}{secx + tanx}dx = ln |secx +tanx | + c $

$ \displaystyle \int cosecx dx = \int \frac{cosecx (cotx -cosecx)}{cotx – cosecx}dx = ln |cotx – cosecx | + c $

Standard Formulae:

$ \displaystyle \int \frac{dx}{\sqrt{x^2 + a^2}} dx = ln |x + \sqrt{x^2 + a^2}| + c $

$ \displaystyle \int \frac{dx}{\sqrt{x^2 – a^2}} dx = ln |x + \sqrt{x^2 – a^2}| + c $

$ \displaystyle \int \frac{dx}{x^2 + a^2} dx = \frac{1}{2a} ln |\frac{x-a}{x+a}| + c $

$ \displaystyle \int \frac{dx}{a^2 – x^2} dx = \frac{1}{2a} ln |\frac{a+x}{a-x}| + c $

$ \displaystyle \int \sqrt{u^2 +a^2} du = \frac{u}{2} \sqrt{u^2 +a^2} + \frac{a^2}{2} ln |u+\sqrt{u^2 + a^2}| + c $

$ \displaystyle \int \sqrt{u^2 – a^2} du = \frac{u}{2} \sqrt{u^2 – a^2} – \frac{a^2}{2} ln |u+\sqrt{u^2 – a^2}| + c $

$ \displaystyle \int \sqrt{a^2 – x^2} du = \frac{x}{2} \sqrt{a^2 – x^2} + \frac{a^2}{2} sin^{-1}\frac{x}{a} + c $

Also Read :

Indefinite Integral : Basic Concepts
Integration by substitution(Direct & Indirect Substitution)
Integration by Derived Substitution
Integration by Parts
Integration by Partial Fractions

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