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:
\[1.\int x^n \ dx= \frac{x^n+1}{n+1} +c\]
\[2.\int 1\ dx = x + c\]
\[3.\int \frac{1}{x}\ dx = log x + c\]
\[4.\int e^x \ dx= e^x +c\]
\[5.\int sin x \ dx= – cos x +c\]
\[6.\int cosx \ dx= sin x +c\]
\[7.\int sec^2x \ dx= tan x +c\]
\[8.\int cosec^2x \ dx= – cot x +c\]
\[9.\int sec x\ tan x\ dx= sec x +c\]
\[10.\int cosec x\ cot x\ dx= – cosec x +c\]
\[11.\int sin ax \ dx= -\frac{1}{a} cos ax +c\]
\[12.\int cos ax \ dx= \frac{1}{a} sin ax +c\]
\[\color {purple} {Example\ 1\ .}\ \color {red}{Evaluate\ :} \int(x^2 -x + 5)\ dx\ \hspace{10cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int(x^2 -x + 5)\ dx = \frac{x^3}{3} – \frac{x^2}{2} + 5x + c\]
\[\color {purple} {Example\ 2\ .}\ \color {red}{Evaluate\ :} \int(\frac{100}{x}+100)\ dx\ \hspace{10cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int(\frac{100}{x}+100)\ dx = 100\int \frac{1}{x}\ dx + 100\int 1\ dx\]
= 100 log x +100 x + c
\[\therefore\ \boxed{\int(\frac{100}{x}+100)\ dx = 100\ log\ x\ +\ 100\ x\ +\ c}\]
\[\color {purple} {Example\ 3\ .}\ \color {red}{Evaluate\ :} \int(x^2 + \frac{3}{x})\ dx\ \hspace{10cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int(x^2 + \frac{3}{x})\ dx = \frac{x^3}{3} + 3log x + c\]
\[\color {purple} {Example\ 4\ .}\ \color {red}{Evaluate\ :} \int(2 + x)^2\ dx\ \hspace{10cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int(2 + x)^2\ dx\ =\int(4+ 4x + x^2)\ dx\]
\[=4\int(1)dx + 4\int(x)dx + \int(x^2) dx\]
\[=4x + 4 \frac{x^2}{2} + \frac{x^3}{3} + c\]
\[=4x + 2 x^2 + \frac{x^3}{3} + c\]
\[\therefore\ \boxed{\int(2 + x)^2\ dx\ =\ 4x + 2 x^2 + \frac{x^3}{3} + c}\]
\[\color {purple} {Example\ 5\ .}\ \color {red}{Evaluate\ :} \int(x + 6 ) ( x + 2)\ dx\ \hspace{10cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int(x + 6 ) ( x + 2)\ dx\]
\[= \int( x^2 +2x + 6x + 12)\ dx\]
\[= \int(x^2 + 8x+ 12)\ dx\]
\[ = \frac{x^3}{3} + 8\frac{x^2}{2} + 12x + c\]
\[ = \frac{x^3}{3} + 4x^2 + 12x + c\]
\[\therefore\ \boxed{\int(x + 6 ) ( x + 2)\ dx\ =\ \frac{x^3}{3} + 4x^2 + 12x + c}\]
\[\color {purple} {Example\ 6\ .}\ \color {red}{Evaluate\ :} \int(3x + 2 ) ( x + 1)\ dx\ \hspace{10cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int(3x + 2 ) ( x + 1)\ dx\]
\[= \int( 3x^2\ +\ 3x\ +\ 2x\ +\ 2)\ dx\]
\[= \int(3x^2\ +\ 5x\ +\ 2)\ dx\]
\[ =\ 3\frac{x^3}{3}\ +\ 5\frac{x^2}{2}\ +\ 2x + c\]
\[ =\ x^3\ +\ \frac{5}{2}\ x^2\ +\ 2x + c\]
\[\therefore\ \boxed{\int(3x + 2 ) ( x + 1)\ dx\ =\ x^3\ +\ \frac{5}{2}\ x^2\ +\ 2x + c}\]
\[\color {purple} {Example\ 7\ .}\ \color {red}{Evaluate\ :} \int(x^2 + x + 1) ( x^2 – x + 1)\ dx\ \hspace{10cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int(x^2 + x + 1) ( x^2 – x + 1)\ dx\]
\[= \int(x^4 – x^3 + x^2 + x^3 – x^2 + x + x^2 – x + 1)\ dx\]
\[= \int(x^4 + x^2 + 1)\ dx\]
\[ = \frac{x^5}{5} + \frac{x^3}{3} + x + c\]
\[\therefore\ \boxed{\int(x^2 + x + 1) ( x^2 – x + 1)\ dx\ =\ \frac{x^5}{5} + \frac{x^3}{3} + x + c}\]
\[\color {purple} {Example\ 8\ .}\ \color {red}{Evaluate\ :}\ \int(2 sin x + 7)\ dx\ \hspace{10cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int(2 sin x + 7)\ dx =2\int sin x\ dx + 7\int 1\ dx\]
\[=- 2 cos x + 7x + c\]
\[\therefore\ \boxed{\int(2 sin x + 7)\ dx\ =\ – 2 cos x + 7x + c}\]
\[\color {purple} {Example\ 9\ .}\ \color {red}{Evaluate\ :} \int(3\ x^2\ -\ 5\ sec^2\ x\ +\ \frac{7}{x})\ dx\ \hspace{10cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int(3\ x^2\ -\ 5\ sec^2\ x\ +\ \frac{7}{x})\ dx\ =\ 3\int x^2\ dx\ -\ 5\ \int Sec^2\ x\ dx\ +\ 7\ \int \frac{1}{x}\ dx\]
\[=\ 3\ \frac{x^3}{3}\ -\ 5\ tan\ x\ +\ 7\ log\ x\ +\ c\]
\[\boxed{\int(3\ x^2\ -\ 5\ sec^2\ x\ +\ \frac{7}{x})\ dx\ =\ x^3\ -\ 5\ tan\ x\ +\ 7\ log\ x\ +\ c}\]
Trigonometry related formulae:
\[1.\ sin^2 x + cos^2 x = 1\]
\[2.\ sin^2 x = \frac{1 – cos 2x}{2}\]
\[3.\ cos^2 x = \frac{1 + cos 2x}{2}\]
\[4.\ tan^2 x = sec^2 x – 1\]
\[5.\ cot^2 x = cosec^2 x – 1\]
\[6.\ sin\ 2x\ =\ 2\ sin\ x\ cos\ x\]
\[7.\ sin 3x = 3 sin x – 4 sin^3 x \]
\[sin^3 x = \frac{1}{4}[3 sin x – sin 3x]\]
\[8.\ cos3x = 4 cos^3 x – 3 cos x\]
\[cos^3 x = \frac{1}{4}[ cos 3 x + 3 cos x]\]
\[9.\ sin ( A+ B) + sin ( A – B) = 2 sin A cos B\]
\[ sin A\ cos B = \frac{1}{2}[sin(A + B) + sin ( A – B)]\]
\[10.\ cos A\ sin B = \frac{1}{2}[sin(A + B)\ -\ sin ( A – B)]\]
\[11. \ cos A\ cos B = \frac{1}{2}[cos (A + B) + cos ( A – B)]\]
\[12. \ sin A\ sin B = \frac{1}{2}[cos (A + B)\ -\ cos ( A – B)]\]
\[\color {purple} {Example\ 10:}\ \color {red}{Evaluate:}\ \int \sqrt{1\ +\ sin\ 2x}\ dx\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int \sqrt{1\ +\ sin\ 2x} \ dx = \int \sqrt{sin^2\ x\ +\ cos^2\ x\ +\ 2\ sin\ x\ cos\ x} \ dx\]
\[= \int \sqrt{(sin\ x\ +\ cos\ x)^2}\ dx\]
\[= \int (sin\ x\ +\ cos\ x)\ dx\]
\[= \int\ sin\ x\ dx\ +\ \int cos\ x\ dx\]
\[=\ -\ cos\ x\ +\ sin\ x\ +\ c\]
\[\therefore\ \boxed{\int \sqrt{1\ +\ sin\ 2x}\ dx\ =\ -\ cos\ x\ +\ sin\ x\ +\ c}\]
\[\color {purple} {Example\ 11:}\ \color {red}{Evaluate:} \int sin^2 x \ dx\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int sin^2 x \ dx = \int (\frac{1 – cos 2x}{2})\ dx\]
\[ = \frac{1}{2}\ [\int 1\ dx\ -\ \int cos 2x]\ dx\]
\[=\frac{1}{2}[ x – \frac{1}{2} sin 2x] + c\]
\[\therefore\ \boxed{\int sin^2 x \ dx\ =\ \frac{1}{2}[ x – \frac{1}{2} sin 2x] + c}\]
\[\color {purple} {Example\ 12:}\ \color {red}{Evaluate:} \int cos^2 x \ dx\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int cos^2 x \ dx = \int (\frac{1 + cos 2x}{2})\ dx\]
\[=\frac{1}{2}[ x + \frac{1}{2} sin 2x] + c\]
\[\therefore\ \boxed{\int cos^2 x \ dx\ =\ -\frac{1}{2}[ x + \frac{1}{2} sin 2x] + c}\]
\[\color {purple} {Example\ 13:}\ \color {red}{Evaluate:} \int tan^2 x \ dx\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int tan^2 x \ dx = \int(sec^2 x – 1)\ dx\]
\[=\int sec^2 x\ dx – \int 1\ dx\]
= tan x – x+ c
\[\therefore\ \boxed{\int tan^2 x \ dx\ =\ tan\ x\ -\ x\ + c}\]
\[\color {purple} {Example\ 14:}\ \color {red}{Evaluate:} \int \frac{sin^2 x}{ 1- cos x} \ dx\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int \frac{sin^2 x}{ 1- cos x} \ dx = \int\frac{1 – cos^2 x}{ 1- cos x} \ dx\]
\[= \int\frac{(1+ cos x)(1 – cos x)}{1- cos x}\ dx\]
\[=\int ( 1 + cos x)\ dx\]
\[= x + sin x +c\]
\[\therefore\ \boxed{\int \frac{sin^2 x}{ 1- cos x} \ dx\ =\ x + sin x +c}\]
\[\color {purple} {Example\ 15:}\ \color {red}{Evaluate:} \int \frac{cos^2 x}{ 1- sin x} \ dx\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int \frac{cos^2 x}{ 1- sin x} \ dx = \int\frac{1 – sin^2 x}{ 1- sin x} \ dx\]
\[= \int\frac{(1+ sin x)(1 – sin x)}{1- sin x}\ dx\]
\[=\int ( 1 + sin x)\ dx\]
= x – cos x + c
\[\therefore\ \boxed{\int \frac{cos^2 x}{ 1- sin x} \ dx\ =\ x\ -\ cos x\ +\ c}\]
\[\color {purple} {Example\ 16:}\ \color {red}{Evaluate:} \int ( sin x + cos x ) ^ 2\ dx\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int ( sin x + cos x ) ^ 2\ = \int ( sin ^2 x + cos ^2 x + 2 sin x cos x)\ dx\]
\[= \int ( 1 + sin 2x)\ dx\]
\[= \int 1\ dx + \int sin 2x\ dx\]
\[= x – \frac{1}{2} cos 2x + c\]
\[\therefore\ \boxed{\int ( sin x + cos x ) ^ 2\ dx\ =\ x – \frac{1}{2} cos 2x + c}\]
\[\color {purple} {Example\ 17:}\ \color {red}{Evaluate:}\ \int ( tan\ x\ +\ cot\ x ) ^ 2\ dx\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int (tan\ x\ +\ cot\ x)^2 \ dx\ =\ \int(tan^2\ x\ +\ cot^2\ x\ +\ 2\ tan\ x\ cot\ x)\ dx\]
\[= \int\ tan^2\ x\ dx\ +\ \int cot^2\ x\ dx\ +\ \int 2\ dx\]
\[= \int\ (sec^2\ x\ -\ 1)dx\ +\ \int (cosec^2\ x\ -\ 1)dx\ +\ 2\int dx\]
\[=\ \ tan\ x\ -\ x\ +\ cot\ x\ -\ x\ +\ 2\ x\ +\ c\]
\[=\ \ tan\ x\ +\ cot\ x\ +\ c\]
\[\therefore\ \boxed{\int(tan\ x\ +\ cot\ x)^2\ dx\ =\ tan\ x\ +\ cot\ x\ +\ c}\]
\[\color {purple} {Example\ 18:}\ \color {red}{Evaluate\ :}\ \int cos^3 x \ dx\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int cos^3 x \ dx = \int \frac{1}{4}[ cos 3 x\ +\ 3 cos x]\ dx\]
\[=\frac{1}{4}[\int cos 3x\ dx – 3\int cos x\ dx]\]
\[=\frac{1}{4}[ \frac{1}{3}\ sin 3x\ -\ 3\ sin\ x] + c\]
\[\therefore\ \boxed{\int cos^3 x \ dx\ =\ \frac{1}{12}\ sin\ 3x\ -\ 3\ sin\ x\ + c}\]
\[\color {purple} {Example\ 19:}\ \color {red}{Evaluate\ :} \int sin 5x\ cos 2x\ dx\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int sin 5x\ cos 2x\ dx = \frac{1}{2}[\int((sin (5x +2x) + sin ( 5x – 2x))\ dx]\]
\[= \frac{1}{2}[\int(sin 7x + sin 3x)\ dx]\]
\[= \frac{1}{2}[\int sin 7x\ dx + \int sin 3x\ dx]\]
\[= \frac{1}{2}[-\frac{cos7x}{7} – \frac{cos3x}{3}]\ +\ c\]
\[\therefore\ \boxed{\int sin 5x\ cos 2x\ dx\ =\ \frac{1}{2}[-\frac{cos7x}{7} – \frac{cos3x}{3}]\ +\ c}\]
\[\color {purple} {Example\ 20:}\ \color {red}{Evaluate\ :} \int sin 3x\ sin x\ dx\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int sin 3x\ sin x\ dx\ =\ -\ \frac{1}{2}[\int((cos (3x\ +\ x)\ -\ cos ( 3x\ -\ x))\ dx]\]
\[=\ -\ \frac{1}{2}[\int(cos\ 4x\ -\ cos\ 2x)\ dx]\]
\[=\ -\ \frac{1}{2}[\int cos\ 4x\ dx\ -\ \int cos\ 2x\ dx]\]
\[=\ – \frac{1}{2}[\frac{sin\ 4x}{4}\ -\ \frac{sin\ 2x}{2}]\ +\ c\]
\[\therefore\ \boxed{\int sin 3x\ sin x\ dx\ =\ – \frac{1}{2}[\frac{sin\ 4x}{4}\ -\ \frac{sin\ 2x}{2}]\ +\ c}\]
Exercise Problems
\[\LARGE{\color {purple} {PART- A}}\]
\[\color {purple} {1\ .}\ \color {red} {Evaluate\ :}\ \int(5x^2\ -\ \frac{2}{x^3}\ +\ \frac{1}{x}\ -\ 3)\ dx\ \hspace{18cm}\]
\[\color {purple} {2\ .}\ \color {red} {Evaluate\ :}\ \int(x^2 \ +\ cos\ x)\ dx\ \hspace{18cm}\]
\[\color {purple} {3\ .}\ \color {red} {Evaluate\ :} \int(x^2 \ +\ Sec^2\ x)\ dx\ \hspace{18cm}\]
\[\color {purple} {4\ .}\ \color {red} {Evaluate\ :} \int cot^2 x \ dx\ \hspace{18cm}\]
\[\LARGE{\color {purple} {PART- B}}\]
\[\color {purple} {5\ .}\ \color {red} {Evaluate\ :} \int(x^2 + x + 1) (x\ +\ 5)\ dx\ \hspace{18cm}\]
\[\color {purple} {6\ .}\ \color {red} {Evaluate\ :} \int \frac{sin^2 x}{ 1\ +\ cos x} \ dx\ \hspace{18cm}\]
\[\color {purple} {7\ .}\ \color {red} {Evaluate\ :} \int \frac{dx}{1\ -\ sin\ x}\ \hspace{18cm}\]
\[\LARGE{\color {purple} {PART- C}}\]
\[\color {purple} {8\ .}\ \color {red} {Evaluate\ :}\ (i)\ \int (\sqrt{x}\ +\ \frac{1}{\sqrt{x}}) \ dx\ \hspace{2cm}\ (ii)\ \int cos\ x\ cos\ 12\ x\ \ dx\ \hspace{8cm}\]
\[\color {purple} {9\ .}\ \color {red} {Evaluate\ :}\ (i)\ \int (x\ -\ 1) (2x\ +\ 3)\ dx\ \hspace{2cm}\ (ii)\ \int 2\ sin\ 3\ x\ cos\ x\ \ dx\ \hspace{8cm}\]
\[\color {purple} {10\ .}\ \color {red} {Evaluate\ :}\ \int sin^3\ x\ dx\ \hspace{18cm}\]
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