# INTEGRATION – DECOMPOSITION METHOD EXERCISE PROBLEMS WITH SOLUTIONS

$\LARGE{\color {purple} {PART- A}}$
$\color {purple} {1\ .}\ \color {red} {Evaluate\ :}\ \int(5x^2\ -\ \frac{2}{x^3}\ +\ \frac{1}{x}\ -\ 3)\ dx\ \hspace{18cm}$
$\color {blue} {Soln\ :}\ \hspace{18cm}$
$\int(5x^2\ -\ \frac{2}{x^3}\ +\ \frac{1}{x}\ -\ 3)\ dx\ =\ 5\int x^2\ dx\ -\ 2\int x^{-3}\ dx\ +\ \int \frac{1}{x}\ dx\ -\ 3\int 1\ dx$
$=\ 5\ \frac{x^3}{3}\ – 2 \frac{x^{-2}}{-2}\ +\ log\ x\ -\ 3\ x\ +\ c$
$=\ 5\ \frac{x^3}{3}\ +\ x^{-2}\ +\ log\ x\ -\ 3\ x\ +\ c$
$\therefore\ \boxed{\int(5x^2\ -\ \frac{2}{x^3}\ +\ \frac{1}{x}\ -\ 3)\ dx\ =\ 5\ \frac{x^3}{3}\ +\ x^{-2}\ +\ log\ x\ -\ 3\ x\ +\ c}$

$\color {purple} {2\ .}\ \color {red} {Evaluate\ :}\ \int(x^2 \ +\ cos\ x)\ dx\ \hspace{18cm}$
$\color {blue} {Soln\ :}\ \hspace{18cm}$
$\int(x^2 \ +\ cos\ x)\ dx\ =\ \int x^2\ dx\ +\ \int cos\ x\ dx$
$=\ \frac{x^3}{3}\ +\ Sin\ x\ +\ c$
$\therefore\ \boxed{\int(x^2 \ +\ cos\ x)\ dx\ =\ \frac{x^3}{3}\ +\ Sin\ x\ +\ c}$

$\color {purple} {3\ .}\ \color {red} {Evaluate\ :} \int(x^2 \ +\ Sec^2\ x)\ dx\ \hspace{18cm}$
$\color {blue} {Soln\ :}\ \hspace{18cm}$
$\int(x^2 \ +\ Sec^2\ x)\ dx\ =\ \int x^2\ dx\ +\ \int Sec^2\ x\ dx$
$=\ \frac{x^3}{3}\ +\ tan\ x\ +\ c$
$\therefore\ \boxed{\int(x^2 \ +\ Sec^2\ x)\ dx\ =\ \frac{x^3}{3}\ +\ tan\ x\ +\ c}$

$\color {purple} {4\ .}\ \color {red} {Evaluate\ :} \int cot^2 x \ dx\ \hspace{18cm}$
$\color {blue} {Soln\ :}\ \hspace{18cm}$
$\int cot^2 x \ dx = \int(cosec^2 x – 1)\ dx$
$=\int cosec^2 x\ dx – \int 1\ dx$
$=\ -\ cot\ x\ -\ x\ +\ c$
$\therefore\ \boxed{\int cot^2 x \ dx\ =\ -\ cot\ x\ -\ x\ +\ c}$

$\LARGE{\color {purple} {PART- B}}$
$\color {purple} {5\ .}\ \color {red} {Evaluate\ :} \int(x^2 + x + 1) (x\ +\ 5)\ dx\ \hspace{18cm}$
$\color {blue} {Soln\ :}\ \hspace{18cm}$
$\int (x^2 + x + 1) (x\ +\ 5)\ dx = \int(x^3\ +\ x^2\ +\ x\ +\ 5\ x^2\ +\ 5\ x\ +\ 5)\ dx$
$=\int (x^3\ +\ 6\ x^2\ +\ 6\ x\ +\ 5)\ dx$
$=\ \frac{x^4}{4}\ +\ 6 \frac{x^3}{3}\ +\ 6\ \frac{x^2}{2}\ +\ 5\ x\ +\ c$
$=\ \frac{x^4}{4}\ +\ 2\ x^3\ +\ 3\ x^2\ +\ 5\ x\ +\ c$
$\therefore\ \boxed{\int (x^2 + x + 1) (x\ +\ 5)\ dx\ =\ \frac{x^4}{4}\ +\ 2\ x^3\ +\ 3\ x^2\ +\ 5\ x\ + c}$

$\color {purple} {6\ .}\ \color {red} {Evaluate\ :} \int \frac{sin^2 x}{ 1\ +\ cos x} \ dx\ \hspace{18cm}$
$\color {blue} {Soln\ :}\ \hspace{18cm}$
$\int \frac{sin^2 x}{ 1\ +\ cos x} \ dx = \int\frac{1 – cos^2 x}{ 1\ +\ cos x} \ dx$
$= \int\frac{(1+ cos x)(1 – cos x)}{1\ +\ cos x}\ dx$
$=\int ( 1\ -\ cos x)\ dx$
$=\ x\ -\ sin x\ +\ c$
$\therefore\ \boxed{\int \frac{sin^2 x}{ 1\ +\ cos x} \ dx\ =\ x\ -\ sin x\ +\ c}$

$\color {purple} {7\ .}\ \color {red} {Evaluate\ :} \int \frac{dx}{1\ -\ sin\ x}\ \hspace{18cm}$
$\color {blue} {Soln\ :}\ \hspace{18cm}$
$\int \frac{dx}{1\ -\ sin\ x} \ dx = \int\frac{dx}{ 1\ -\ sin\ x} \ ×\ \frac{1\ +\ sin\ x}{1\ -\ sin\ x}$
$= \int\frac{1\ +\ sin\ x}{1\ -\ sin^2 x}\ dx$
$= \int\frac{1\ +\ sin\ x}{cos^2 x}\ dx$
$= \int\frac{1}{cos^2 x}\ dx\ +\ \int\frac{sin\ x}{cos^2 x}\ dx$
$= \int\ sec^2 x\ dx\ +\ \int tan\ x\ sec\ x\ dx$
$=\ tan\ x\ +\ sec\ x\ +\ c$
$\therefore\ \boxed{\int \frac{dx}{ 1\ -\ sin\ x}\ =\ tan\ x\ +\ sec\ x\ +\ c}$

$\LARGE{\color {purple} {PART- C}}$
$\color {purple} {8\ .}\ \color {red} {Evaluate\ :}\ (i)\ \int (\sqrt{x}\ +\ \frac{1}{\sqrt{x}}) \ dx\ \hspace{2cm}\ (ii)\ \int cos\ x\ cos\ 12\ x\ \ dx\ \hspace{8cm}$
$\color {blue} {Soln\ :}\ \hspace{18cm}$
$(i)\ \int (\sqrt{x}\ +\ \frac{1}{\sqrt{x}}) \ dx\ =\ \int\ \sqrt{x}\ dx\ +\ \int \frac{1}{\sqrt{x}}\ dx$
$= \int\ x^{\frac{1}{2}}\ dx\ +\ \int x^{\frac{-1}{2}}\ dx$
$= \ \frac{x^{\frac{3}{2}}}{{\frac{3}{2}}}\ +\ \frac{x^{\frac{1}{2}}}{{\frac{1}{2}}}\ +\ c$
$= \ \frac{2}{3}\ x^{\frac{3}{2}}\ +\ 2\ x^{\frac{1}{2}}\ +\ c$
$\therefore\ \boxed{\int (\sqrt{x}\ +\ \frac{1}{\sqrt{x}})\ dx\ =\ \frac{2}{3}\ x^{\frac{3}{2}}\ +\ 2\ x^{\frac{1}{2}}\ +\ c}$

$(ii)\ \int cos\ x\ cos\ 12\ x\ dx\ =\ \frac{1}{2}[\int((cos\ (x\ +\ 12x)\ +\ cos(x\ -\ 12x))\ dx]$
$= \frac{1}{2}[\int(cos\ 13\ x\ +\ cos\ (-11x))\ dx]$
$= \frac{1}{2}[\int cos\ 13\ x\ dx\ +\ \int cos\ 11\ x\ dx]$
$= \frac{1}{2}[\frac{sin\ 13\ x}{13}\ +\ \frac{sin\ 11\ x}{11}]\ +\ c$
$\therefore\ \boxed{\int cos\ x\ cos\ 12\ x\ dx\ =\ \frac{1}{2}[\frac{sin\ 13\ x}{13}\ +\ \frac{sin\ 11\ x}{11}]\ +\ c}$

$\color {purple} {9\ .}\ \color {red} {Evaluate\ :}\ (i)\ \int (x\ -\ 1) (2x\ +\ 3)\ dx\ \hspace{2cm}\ (ii)\ \int 2\ sin\ 3\ x\ cos\ x\ \ dx\ \hspace{8cm}$
$\color {blue} {Soln\ :}\ \hspace{18cm}$
$(i)\ \int (x\ -\ 1) (2x\ +\ 3) \ dx\ =\ \int(2\ x^2\ +\ 3\ x\ -\ 2\ x\ -\ 3)\ dx$
$= \int(2\ x^2\ +\ x\ -\ 3)\ dx$
$=\ 2 \frac{x^3}{3}\ +\ \frac{x^2}{2}\ -\ 3\ x + c$
$\therefore\ \boxed{\int(x\ -\ 1) ( 2x\ +\ 3)\ dx\ =\ 2 \frac{x^3}{3}\ +\ \frac{x^2}{2}\ -\ 3\ x + c}$

$(ii)\ \int 2\ sin 3x\ cos x\ dx\ =\ \int((sin (3x\ +\ x)\ +\ sin ( 3x\ -\ x))\ dx]$
$=\ \int(sin\ 4x\ +\ sin\ 2x)\ dx$
$=\ \int sin 4x\ dx\ +\ \int sin 2x\ dx$
$=\ -\frac{cos\ 4x}{4}\ -\ \frac{cos\ 2x}{2}\ +\ c$
$\therefore\ \boxed{\int 2\ sin 3x\ cos x\ dx\ =\ -\frac{cos\ 4x}{4}\ -\ \frac{cos\ 2x}{2}\ +\ c}$

$\color {purple} {10.}\ \color {red}{Evaluate\ :} \int sin 7x\ sin 3x\ dx\ \hspace{15cm}$
$\color {purple} {11\ .}\ \color {red} {Evaluate\ :}\ \int sin^3\ x\ dx\ \hspace{18cm}$
$\color {blue} {Soln\ :}\ \hspace{18cm}$
$\int sin^3\ x dx\ \hspace{10cm}$
$W.\ K.\ T\ sin\ 3\ x\ =\ 3\ sin\ x\ -\ 4\ sin^3\ x$
$sin^3\ x\ =\ \frac{1}{4}[3\ sin\ x\ -\ sin\ 3\ x]$
$\int sin^3\ x\ dx\ =\ \int \frac{1}{4}[3\ sin\ x\ -\ sin\ 3\ x]\ dx$
$=\frac{1}{4}[3\int sin\ x\ dx – \int sin\ 3x\ dx]$
$=\frac{1}{4}[ -\ 3\ cos\ x\ +\ \frac{cos\ 3x}{3}]\ + c$
$\therefore\ \boxed{\int sin^3 x \ dx\ =\ \frac{1}{4}[-\ 3\ cos\ x\ +\ \frac{cos\ 3x}{3}] + c}$