4.3 STANDARD INTEGRALS

\[Integrals\ of\ the\ form\ \int \frac{dx}{a^2 \pm x^2}, \int \frac{dx}{x^2-a^2}\ ,\ \int \frac{dx}{\sqrt{a^2 – x^2}}\]
\[\int {\sqrt{a^2 - x^2}}\ dx\ and\ \int {\sqrt{x^2 \pm a^2}}\ dx\]

List of  Formulae:

\[1.\ \int \frac{dx}{a^2 + x^2} = \frac{1}{a}\ {tan}^{-1}(\frac{x}{a}) +c\ (or)\ \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\ {tan}^{-1} (\frac{x}{a}) +c\]
\[2.\ \int \frac{dx}{a^2 - x^2} = \frac{1}{2a}\ log\ (\frac{a + x}{a - x}) + c\]
\[3.\ \int \frac{dx}{x^2 - a^2} = \frac{1}{2a}\ log\ (\frac{x - a}{x + a}) + c\]
\[4.\ \int \frac{dx}{{\sqrt{a^2 - x^2}}} = {sin}^{-1}(\frac{x}{a})\ + c\]
\[5.\ \int {\sqrt{a^2 - x^2}}\ dx = \frac{x}{2}\ {\sqrt{a^2 - x^2}}\ + \frac{a^2}{2} {sin}^{-1}(\frac{x}{a})\ + c\]
\[6.\ \int {\sqrt{x^2 + a^2}}\ dx = \frac{x}{2}\ {\sqrt{x^2 + a^2}}\ + \frac{a^2}{2} {sin\ h}^{-1}(\frac{x}{a})\ + c\]
\[7.\ \int {\sqrt{x^2 - a^2}}\ dx = \frac{x}{2}\ {\sqrt{x^2 - a^2}}\ - \frac{a^2}{2} {cos\ h}^{-1}(\frac{x}{a})\ + c\]

Example :

\[Evaluate:\ \int \frac{dx}{x^2 + 16}\]

Soln:

\[W.K.T\ \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\ {tan}^{-1} (\frac{x}{a}) +c\]
\[\int \frac{dx}{x^2 + 16} = \int \frac{dx}{x^2 + 4^2}\]
\[ = \frac{1}{4}\ {tan}^{-1} (\frac{x}{4}) +c\]

Example :

\[Evaluate:\ \int \frac{dx}{4x^2 + 9}\]

Soln:

\[\int \frac{dx}{4x^2 + 9} = \frac{1}{4}\ \int \frac{dx}{x^2 + \frac{9}{4}} = \frac{1}{4}\ \int \frac{dx}{x^2 + (\frac{3}{2})^2}\]
\[W.K.T\ \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\ {tan}^{-1} (\frac{x}{a}) +c\]
\[\int \frac{dx}{4x^2 + 9} = \frac{1}{4}\ × \frac{2}{3}\ {tan}^{-1} (\frac{x}{\frac{3}{2}}) +c\]
\[ = \frac{1}{6}\ {tan}^{-1} (\frac{2}{3}) +c\]

Example :

\[Evaluate:\ \int \frac{dx}{4 + 9 x^2}\]

Soln:

\[\int \frac{dx}{4 + 9 x^2} = \int \frac{dx}{9 x^2+ 4} = \frac{1}{9}\ \int \frac{dx}{x^2 + \frac{4}{9}} = \frac{1}{9}\ \int \frac{dx}{x^2 + (\frac{2}{3})^2}\]
\[W.K.T\ \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\ {tan}^{-1} (\frac{x}{a}) +c\]
\[\int \frac{dx}{4 + 9 x^2} = \frac{1}{9}\ × \frac{3}{2}\ {tan}^{-1} (\frac{x}{\frac{2}{3}}) +c\]
\[ = \frac{1}{6}\ {tan}^{-1} (\frac{3x}{2}) +c\]

Example :

\[Evaluate:\ \int \frac{dx}{(3x + 2)^2 + 16}\]

Soln:

Put      u  = 3x + 2

\[\frac{du}{dx}= \ 3\]
\[dx = \frac{1}{3}\ du\]
\[\int \frac{dx}{(3x + 2)^2 + 16} = \frac{1}{3}\ \int \frac{du}{u^2 + 4^2}\]
\[W.K.T\ \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\ {tan}^{-1} (\frac{x}{a}) +c\]
\[\frac{1}{3}\ \int \frac{du}{u^2 + 4^2} = \frac{1}{3}\ × \frac{1}{4}\ {tan}^{-1} (\frac{u}{4}) +c\]
\[ = \frac{1}{12}\ {tan}^{-1} (\frac{3x + 2}{4}) +c\]
\[\int \frac{dx}{(3x + 2)^2 + 16} = \frac{1}{12}\ {tan}^{-1} (\frac{3x + 2}{4}) +c\]

Example :

\[Evaluate:\ \int \frac{dx}{9 - (3x - 2)^2}\]

Soln:

Put      u  = 3x - 2

\[\frac{du}{dx}= \ 3\]
\[dx = \frac{1}{3}\ du\]
\[\int \frac{dx}{9 - (3x - 2)^2}= \frac{1}{3}\ \int \frac{du}{3^2 - u^2}\]
\[W.K.T\ \int \frac{dx}{a^2 - x^2} = \frac{1}{2a}\ log\ (\frac{a + x}{a - x}) + c\]
\[\frac{1}{3}\ \int \frac{du}{3^2 - u^2} = \frac{1}{3}\ [ \frac{1}{2 × 3 }\ log\ (\frac{3 + u}{3 - u})] +c\]
\[=\frac{1}{18}\ [log\ (\frac{3 + (3x - 2)}{3 - (3x - 2)})] +c\]
\[=\frac{1}{18}\ [log\ (\frac{3 + 3x - 2}{3 - 3x + 2})] +c\]
\[=\frac{1}{18}\ [log\ (\frac{1 + 3x }{5 - 3x })] +c\]

Example :

\[Evaluate:\ \int \frac{dx}{{\sqrt{9 - x^2}}}\]

Soln:

\[\int \frac{dx}{{\sqrt{9 - x^2}}} = \int \frac{dx}{{\sqrt{3^2 - x^2}}} \]
\[W.K.T\ \int \frac{dx}{{\sqrt{a^2 - x^2}}} = {sin}^{-1}(\frac{x}{a})\ + c\]
\[\int \frac{dx}{{\sqrt{9 - x^2}}} = {sin}^{-1}(\frac{x}{3})\ + c\]

Example :

\[Evaluate:\ \int \frac{dx}{{\sqrt{5 - 4x^2}}}\]

Soln:

\[\int \frac{dx}{{\sqrt{5 - 4x^2}}} = \frac{1}{2}\ \int \frac{dx}{{\sqrt{\frac{5}{4} - x^2}}}\]
\[= \frac{1}{2}\ \int \frac{dx}{{\sqrt{(\frac{\sqrt{5}}{2})^2 - x^2}}}\]
\[W.K.T\ \int \frac{dx}{{\sqrt{a^2 - x^2}}} = {sin}^{-1}(\frac{x}{a})\ + c\]
\[\int \frac{dx}{{\sqrt{5 - 4x^2}}} = \frac{1}{2}\ {sin}^{-1}(\frac{x}{\frac{\sqrt{5}}{2}})\ + c\]
\[= \frac{1}{2}\ {sin}^{-1}(\frac{2x}{\sqrt{5}})\ + c\]
\[\int \frac{dx}{{\sqrt{5 - 4x^2}}} = \frac{1}{2}\ {sin}^{-1}(\frac{2x}{\sqrt{5}})\ + c\]

Example :

\[Evaluate:\ \int {\sqrt{9x^2 +16}}\ dx\]

Soln:

\[Evaluate:\ \int {\sqrt{9x^2 +16}}\ dx =3 \int {\sqrt{x^2 +\frac{16}{9}}}\ dx = 3 \int {\sqrt{x^2 + (\frac{4}{3}}})^2\ dx\]
\[\int {\sqrt{x^2 + a^2}}\ dx = \frac{x}{2}\ {\sqrt{x^2 + a^2}}\ + \frac{a^2}{2} {sin\ h}^{-1}(\frac{x}{a})\ + c\]
\[\int {\sqrt{9x^2 +16}}\ dx = 3[\frac{x}{2}\ {\sqrt{x^2 + \frac{16}{9}}}\ + \frac{16}{18} {sin\ h}^{-1}(\frac{x}{\frac{4}{3}})]\ + c\]
\[ = 3[\frac{x}{2}\ {\sqrt{x^2 + \frac{16}{9}}}\ + \frac{8}{9} {sin\ h}^{-1}(\frac{3x}{4})]\ + c\]

4.2 INTEGRATION BY SUBSTITUTION

So far we have dealt with functions, either directly integrable using integration formula (or) integrable after decomposing the given functions into sums & differences.

\[But\ there\ are\ functions\ like\ \frac{sin(log x)}{x},\ \frac{2x + 3}{x^2 + 3x + 5}\]

which cannot be decomposed into sums (or) differences of simple functions. In these cases, using proper substitution, we shall reduce the given form into standard form, which can be integrated using basic integration formula.

When the integrand (the function to be integrated) is either in multiplication or in division form and if the derivative of one full or meaningful part of the function is equal to the other function then the integration can be evaluated using substitution method as given in the following examples.

\[1.\ \int \frac{sin(log\ x)}{x}\ dx= \int sin (log\ x) \frac{1}{x}\ dx\]
\[Here\ \frac{d}{dx} (log\ x) = \frac{1}{x}\]

The above integration can be evaluated by taking  u = log x.

\[2.\ \int \frac{2x + 3}{x^2 + 3x + 5}\ dx\]
\[since\ \frac{d}{dx} ( x^2 + 3x + 5)\ is\ 2x + 3\]

it can be integrated by  taking  u  =x2 + 3x + 5.

Integrals of the form

\[\int[f(x)]^n\ f^!(x)\ dx,\ n\ \neq\ -1,\]
\[\int\ \frac{f^!(x)}{f(x)}\ dx\]

and

\[\int[F(f(x))]\ f^!(x)\ dx\]

are all, more or less of the same type and the use of substitution u = f(x) will reduce the given function to simple standard form which can be integrated using integration formulae.

Example:

\[Evaluate\ :\ \int\ (x^2 + x + 1)^5\ (2x + 1) dx\]

Soln:

\[put\ u =x^2 + x + 1\]
\[\frac{du}{dx}= \ 2x + 1\]
\[\int\ (x^2 + x + 1)^5\ (2x + 1) dx= \int\ u^5\ du\]
\[=\frac{u^6}{6} + c\]
\[=\frac{(x^2 + x + 1)^6}{6} + c\]

Example:

\[Evaluate\ :\ \int\ \frac{2x}{1 + x^2}\ dx\]

Soln:

\[put\ u\ =\ 1 + x^2\]
\[\frac{du}{dx}= \ 2x\]

du  = 2 dx

\[\int\ \frac{2x}{1 + x^2}\ dx= \int\ \frac{du}{u}\]

=  log u + c

\[= log(1 + x^2) + c\]
\[\int\ \frac{2x}{1 + x^2}\ dx= log(1 + x^2) + c\]

Example:

\[Evaluate\ :\ \int\ \frac{dx}{x\ log x}\]

Soln:

\[\int\ \frac{dx}{x\ log x} = \int\ \frac{1}{log x}\ \frac{1}{x}\ dx\]

Put      u  =     log x

\[\frac{du}{dx}=\frac{1}{x}\]
\[du = \frac{1}{x}\ dx\]
\[\int\ \frac{dx}{x\ log x}\ dx= \int\ \frac{du}{u}\]

=  log u + c

=  log ( log x ) + c

\[\int\ \frac{dx}{x\ log x}\ dx= log ( log x) + c\]

Example:

\[Evaluate\ :\ \int\ \frac{2x + 1}{ x^2 + x + 1}\ dx\]

Soln:

\[put\ u\ =\ x^2 + x + 1\]
\[\frac{du}{dx}= \ 2x + 1\]

du  = (2x + 1) dx

\[\int\ \frac{2x+ 1}{x^2 + x + 1}\ dx= \int\ \frac{du}{u}\]

=  log u + c

\[=log( x^2 + x + 1) + c\]

Example:

\[Evaluate\ :\ \int\ \frac{sec^2(log x)}{ x}\ dx\]

Soln:

   Put    u  = log x

\[\frac{du}{dx} = \frac{1}{x}\]
\[du = \frac{1}{x}\ dx\]
\[\int\ \frac{sec^2(log x)}{ x}\ dx = \int sec^2\ du\]

=  tan u  + c

=   tan ( log x ) + c

Example:

\[Evaluate\ :\ \int\ sin^3x\ cos x dx\]

Soln:

   Put    u  =  sin x

\[\frac{du}{dx} = cos x\]

du  =  cos x  dx

\[\int\ sin^3x\ cos x dx = \int u^3\ du\]
\[=\frac{u^4}{4} + c\]
\[=\frac{sin^4 x}{4} + c\]

Example:

\[Evaluate\ :\ \int\ \frac{2ax + b}{{\sqrt{(ax^2 + bx + c)}}}\ dx\]

Soln:

\[put\ u\ =\ ax^2 + bx + c\]
\[\frac{du}{dx}= \ 2ax + b\]

du  = (2ax + b) dx

\[\int\ \frac{2ax + b}{{\sqrt{(ax^2 + bx + c)}}}\ dx= \int\ \frac{du}{\sqrt{u}}\]
\[=\frac{u^{\frac{-1}{2} + 1}}{\frac{-1}{2} + 1} + c\]
\[=\frac{u^{\frac{1}{2}}}{\frac{1}{2}} + c\]
\[=2u^{\frac{1}{2}} + c\]
\[=2(ax^2+bx+c)^{\frac{1}{2}}\]

4.1 INTEGRATION – DECOMPOSITION METHOD

Sir Sardar Vallabhai Patel, called the Iron Man of India integrated several princely states together while forming our country Indian Nation after independence. Like that in Maths while finding area under a curve through integration, the area under the curve is divided into smaller rectangles and then integrating (i.e) summing of all the area of rectangles together. So, integration means of summation of very minute things of the same kind.

Integration as the reverse of differentiation:

            Integration can also be introduced in another way, called integration as the reverse of differentiation

            Ex:       Suppose we differentiate the function

\[y = x^4 ⇒ \frac{dy}{dx} = 4x^3\]
\[\int 4x^3\ dx\ = 4\frac{x^4}{4} = x^4\]

The symbol for integration is ∫, known as integral sign. Along with the integral sign there is a term dx which must always be written and which indicates the name of the variable involved, in this case ‘x’. Technically integrals of this sort are called indefinite Integrals.

List of  Formulae:

\[1.\int x^n \ dx= \frac{x^n+1}{n+1} +c\]
\[2.\int 1\ dx = x + c\]
\[3.\int \frac{1}{x}\ dx = log x + c\]
\[4.\int e^x \ dx= e^x +c\]
\[5.\int sin x \ dx= – cos x +c\]
\[6.\int cosx \ dx= sin x +c\]
\[7.\int sec^2x \ dx= tan x +c\]
\[8.\int cosec^2x \ dx= – cot x +c\]
\[9.\int sec x\ tan x\ dx= sec x +c\]
\[10.\int cosec x\ cot x\ dx= – cosec x +c\]
\[11.\int sin ax \ dx= -\frac{1}{a} cos ax +c\]
\[12.\int cos ax \ dx= \frac{1}{a} sin ax +c\]

Example

\[Evaluate: \int(x^2 -x -1)\ dx\]

Soln:

\[\int(x^2 -x -1)\ dx = \frac{x^3}{3} – \frac{x^2}{2} – x + c\]

Example

\[Evaluate: \int(\frac{100}{x}+100)\ dx\]

Soln:

\[\int(\frac{100}{x}+100)\ dx = 100\int \frac{1}{x}\ dx + 100\int 1\ dx\]

=  100 log x +100 x + c

Example:

\[Evaluate: \int(x^2 + \frac{3}{x})\ dx\]

Soln:

\[\int(x^2 + \frac{3}{x})\ dx = \frac{x^3}{3} + 3log x + c\]

Example:

\[Evaluate: \int(x^2 + x + 1) ( x^2 – x + 1)\ dx\]

Soln:

\[\int(x^2 + x + 1) ( x^2 – x + 1)\ dx\]
\[= \int(x^4 – x^3 + x^2 + x^3 – x^2 + x + x^2 – x + 1)\ dx\]
\[= \int(x^4 + x^2 + 1)\ dx\]
\[ = \frac{x^5}{5} + \frac{x^3}{3} + x + c\]

Example:

\[Evaluate: \int(2 sin x + 7)\ dx\]

Soln:

\[\int(2 sin x + 7)\ dx =2\int sin x\ dx + 7\int 1\ dx\]
\[=- 2 cos x + 7x + c\]

Trigonometry related formulae:

\[1.\ sin^2 x + cos^2 x = 1\]
\[2.\ sin^2 x = \frac{1 – cos 2x}{2}\]
\[3.\ cos^2 x = \frac{1 + cos 2x}{2}\]
\[4.\ tan^2 x = sec^2 x – 1\]
\[5.\ cot^2 x = cosec^2 x – 1\]
\[6.\ sin 3x = 3 sin x – 4 sin^3 x \]
\[sin^3 x = \frac{1}{4}[3 sin x – sin 3x]\]
\[7.\ cos3x = 4 cos^3 x – 3 cos x\]
\[cos^3 x = \frac{1}{4}[ cos 3 x – 3 cos x]\]
\[8.\ sin ( A+ B) + sin ( A – B) = 2 sin A cos B\]
\[ sin A cos B = \frac{1}{2}[sin(A + B) + sin ( A – B)]\]
\[9. \ cos A cos B = \frac{1}{2}[cos (A + B) + cos ( A – B)]\]

Example:

\[Evaluate: \int cos^2 x \ dx\]

Soln:

\[\int cos^2 x \ dx = \int (\frac{1 + cos 2x}{2})\ dx\]
\[=\frac{1}{2}[ x + \frac{1}{2} sin 2x] + c\]

Example:

\[Evaluate: \int tan^2 x \ dx\]

Soln:

\[\int tan^2 x \ dx = \int(sec^2 x – 1)\ dx\]
\[=\int sec^2 x\ dx – \int 1\ dx\]

=     tan x  –  x+ c

Example:

\[Evaluate: \int \frac{cos^2 x}{ 1- sin x} \ dx\]

Soln:

\[\int \frac{cos^2 x}{ 1- sin x} \ dx = \int\frac{1 – sin^2 x}{ 1- sin x} \ dx\]
\[= \int\frac{(1+ sin x)(1 – sin x)}{1- sin x}\ dx\]
\[=\int ( 1 + sin x)\ dx\]

=  x  – cos x  + c

Example:

\[Evaluate: \int ( sin x + cos x ) ^ 2\ dx\]

Soln:

\[\int ( sin x + cos x ) ^ 2\ = \int ( sin ^2 x + cos ^2 x + 2 sin x cos x)\ dx\]
\[= \int ( 1 + sin 2x)\ dx\]
\[= \int 1\ dx + \int sin 2x\ dx\]
\[= x – \frac{1}{2} cos 2x + c\]

Example:

\[Evaluate: \int cos^3 x \ dx\]

Soln:

\[\int cos^3 x \ dx = \int \frac{1}{4}[ cos 3 x – 3 cos x]\ dx\]
\[=\frac{1}{4}[\int cos 3x\ dx – 3\int cos x\ dx]\]
\[=\frac{1}{4}[ 3sin 3x – 3 cos x] + c\]

Example:

\[Evaluate: \int sin 5x\ cos 2x\ dx\]

Soln:

\[\int sin 5x\ cos 2x\ dx = \frac{1}{2}[\int((sin (5x +2x) + sin ( 5x – 2x))\ dx]\]
\[= \frac{1}{2}[\int(sin 7x + sin 3x)\ dx]\]
\[= \frac{1}{2}[\int sin 7x\ dx + \int sin 3x\ dx]\]
\[= \frac{1}{2}[-\frac{cos7x}{7} – \frac{cos3x}{3}]\]