INTEGRATION – DECOMPOSITION METHOD EXERCISE PROBLEMS WITH SOLUTIONS

\[\underline{PART\ -\ A}\]
\[1.\ Evaluate: \int(5x^2\ -\ \frac{2}{x^3}\ +\ \frac{1}{x}\ -\ 3)\ dx\ \hspace{18cm}\]

Soln:

\[\int(5x^2\ -\ \frac{2}{x^3}\ +\ \frac{1}{x}\ -\ 3)\ dx\ =\ 5\int x^2\ dx\ -\ 2\int x^{-3}\ dx\ +\ \int \frac{1}{x}\ dx\ -\ 3\int 1\ dx\]
\[=\ 5\ \frac{x^3}{3}\ – 2 \frac{x^{-2}}{-2}\ +\ log\ x\ -\ 3\ x\ +\ c\]
\[=\ 5\ \frac{x^3}{3}\ +\ x^{-2}\ +\ log\ x\ -\ 3\ x\ +\ c\]
\[\therefore\ \boxed{\int(5x^2\ -\ \frac{2}{x^3}\ +\ \frac{1}{x}\ -\ 3)\ dx\ =\ 5\ \frac{x^3}{3}\ +\ x^{-2}\ +\ log\ x\ -\ 3\ x\ +\ c}\]
\[2\ Evaluate: \int(x^2 \ +\ cos\ x)\ dx\ \hspace{18cm}\]

Soln

\[\int(x^2 \ +\ cos\ x)\ dx\ =\ \int x^2\ dx\ +\ \int cos\ x\ dx\]
\[=\ \frac{x^3}{3}\ +\ Sin\ x\ +\ c\]
\[\therefore\ \boxed{\int(x^2 \ +\ cos\ x)\ dx\ =\ \frac{x^3}{3}\ +\ Sin\ x\ +\ c}\]
\[3\ Evaluate: \int(x^2 \ +\ Sec^2\ x)\ dx\ \hspace{18cm}\]

Soln

\[\int(x^2 \ +\ Sec^2\ x)\ dx\ =\ \int x^2\ dx\ +\ \int Sec^2\ x\ dx\]
\[=\ \frac{x^3}{3}\ +\ tan\ x\ +\ c\]
\[\therefore\ \boxed{\int(x^2 \ +\ Sec^2\ x)\ dx\ =\ \frac{x^3}{3}\ +\ tan\ x\ +\ c}\]
\[4.\ Evaluate: \int cot^2 x \ dx\ \hspace{18cm}\]

Soln

\[\int cot^2 x \ dx = \int(cosec^2 x – 1)\ dx\]
\[=\int cosec^2 x\ dx – \int 1\ dx\]
\[=\ -\ cot\ x\ -\ x\ +\ c\]
\[\therefore\ \boxed{\int cot^2 x \ dx\ =\ -\ cot\ x\ -\ x\ +\ c}\]
\[\underline{PART\ -\ B}\]
\[5.\ Evaluate: \int(x^2 + x + 1) (x\ +\ 5)\ dx\ \hspace{18cm}\]

Soln

\[\int (x^2 + x + 1) (x\ +\ 5)\ dx = \int(x^3\ +\ x^2\ +\ x\ +\ 5\ x^2\ +\ 5\ x\ +\ 5)\ dx\]
\[=\int (x^3\ +\ 6\ x^2\ +\ 6\ x\ +\ 5)\ dx\]
\[=\ \frac{x^4}{4}\ +\ 6 \frac{x^3}{3}\ +\ 6\ \frac{x^2}{2}\ +\ 5\ x\ +\ c\]
\[=\ \frac{x^4}{4}\ +\ 2\ x^3\ +\ 3\ x^2\ +\ 5\ x\ +\ c\]
\[\therefore\ \boxed{\int (x^2 + x + 1) (x\ +\ 5)\ dx\ =\ \frac{x^4}{4}\ +\ 2\ x^3\ +\ 3\ x^2\ +\ 5\ x\ + c}\]
\[6.\ Evaluate: \int \frac{sin^2 x}{ 1\ +\ cos x} \ dx\ \hspace{18cm}\]

Soln

\[\int \frac{sin^2 x}{ 1\ +\ cos x} \ dx = \int\frac{1 – cos^2 x}{ 1\ +\ cos x} \ dx\]
\[= \int\frac{(1+ cos x)(1 – cos x)}{1\ +\ cos x}\ dx\]
\[=\int ( 1\ -\ cos x)\ dx\]
\[=\ x\ -\ sin x\ +\ c\]
\[\therefore\ \boxed{\int \frac{sin^2 x}{ 1\ +\ cos x} \ dx\ =\ x\ -\ sin x\ +\ c}\]
\[7.\ Evaluate: \int \frac{dx}{1\ -\ sin\ x}\ \hspace{18cm}\]

Soln

\[\int \frac{dx}{1\ -\ sin\ x} \ dx = \int\frac{dx}{ 1\ -\ sin\ x} \ ×\ \frac{1\ +\ sin\ x}{1\ -\ sin\ x}\]
\[= \int\frac{1\ +\ sin\ x}{1\ -\ sin^2 x}\ dx\]
\[= \int\frac{1\ +\ sin\ x}{cos^2 x}\ dx\]
\[= \int\frac{1}{cos^2 x}\ dx\ +\ \int\frac{sin\ x}{cos^2 x}\ dx\]
\[= \int\ sec^2 x\ dx\ +\ \int tan\ x\ sec\ x\ dx\]
\[=\ tan\ x\ +\ sec\ x\ +\ c\]
\[\therefore\ \boxed{\int \frac{dx}{ 1\ -\ sin\ x}\ =\ tan\ x\ +\ sec\ x\ +\ c}\]
\[8.\ Evaluate: \int \sqrt{1\ +\ sin\ 2x}\ dx\ \hspace{18cm}\]

Soln

\[\int \sqrt{1\ +\ sin\ 2x} \ dx = \int \sqrt{sin^2\ x\ +\ cos^2\ x\ +\ 2\ sin\ x\ cos\ x} \ dx\]
\[= \int \sqrt{(sin\ x\ +\ cos\ x)^2}\ dx\]
\[= \int (sin\ x\ +\ cos\ x)\ dx\]
\[= \int\ sin\ x\ dx\ +\ \int cos\ x\ dx\]
\[=\ -\ cos\ x\ +\ sin\ x\ +\ c\]
\[\therefore\ \boxed{\int \sqrt{1\ +\ sin\ 2x}\ dx\ =\ -\ cos\ x\ +\ sin\ x\ +\ c}\]
\[\underline{PART\ -\ C}\]
\[9.\ Evaluate:\ (i)\ \int (\sqrt{x}\ +\ \frac{1}{\sqrt{x}}) \ dx\ \hspace{2cm}\ (ii)\ \int cos\ x\ cos\ 12\ x\ \ dx\ \hspace{8cm}\]

Soln

\[(i)\ \int (\sqrt{x}\ +\ \frac{1}{\sqrt{x}}) \ dx\ =\ \int\ \sqrt{x}\ dx\ +\ \int \frac{1}{\sqrt{x}}\ dx\]
\[= \int\ x^{\frac{1}{2}}\ dx\ +\ \int x^{\frac{-1}{2}}\ dx\]
\[= \ \frac{x^{\frac{3}{2}}}{{\frac{3}{2}}}\ +\ \frac{x^{\frac{1}{2}}}{{\frac{1}{2}}}\ +\ c\]
\[= \ \frac{2}{3}\ x^{\frac{3}{2}}\ +\ 2\ x^{\frac{1}{2}}\ +\ c\]
\[\therefore\ \boxed{\int (\sqrt{x}\ +\ \frac{1}{\sqrt{x}})\ dx\ =\ \frac{2}{3}\ x^{\frac{3}{2}}\ +\ 2\ x^{\frac{1}{2}}\ +\ c}\]
\[(ii)\ \int cos\ x\ cos\ 12\ x\ dx\ =\ \frac{1}{2}[\int((cos\ (x\ +\ 12x)\ +\ cos(x\ -\ 12x))\ dx]\]
\[= \frac{1}{2}[\int(cos\ 13\ x\ +\ cos\ (-11x))\ dx]\]
\[= \frac{1}{2}[\int cos\ 13\ x\ dx\ -\ \int cos\ 11\ x\ dx]\]
\[= \frac{1}{2}[\frac{sin\ 13\ x}{13}\ -\ \frac{sin\ 11\ x}{11}]\ +\ c\]
\[\therefore\ \boxed{\int cos\ x\ cos\ 12\ x\ dx\ =\ \frac{1}{2}[\frac{sin\ 13\ x}{13}\ -\ \frac{sin\ 11\ x}{11}]\ +\ c}\]
\[10.\ Evaluate:\ (i)\ \int (x\ -\ 1) (2x\ +\ 3)\ dx\ \hspace{2cm}\ (ii)\ \int 2\ sin\ 3\ x\ cos\ x\ \ dx\ \hspace{8cm}\]

Soln

\[(i)\ \int (x\ -\ 1) (2x\ +\ 3) \ dx\ =\ \int(2\ x^2\ +\ 3\ x\ -\ 2\ x\ -\ 3)\ dx\]
\[= \int(2\ x^2\ +\ x\ -\ 3)\ dx\]
\[ =\ 2 \frac{x^3}{3}\ +\ \frac{x^2}{2}\ -\ 3\ x + c\]
\[\therefore\ \boxed{\int(x\ -\ 1) ( 2x\ +\ 3)\ dx\ =\ 2 \frac{x^3}{3}\ +\ \frac{x^2}{2}\ -\ 3\ x + c}\]
\[(ii)\ \int 2\ sin 3x\ cos x\ dx\ =\ \int((sin (3x\ +\ x)\ +\ sin ( 3x\ -\ x))\ dx]\]
\[=\ \int(sin\ 4x\ +\ sin\ 2x)\ dx\]
\[=\ \int sin 4x\ dx\ +\ \int sin 2x\ dx\]
\[=\ -\frac{cos\ 4x}{4}\ -\ \frac{cos\ 2x}{2}\ +\ c\]
\[\therefore\ \boxed{\int 2\ sin 3x\ cos x\ dx\ =\ -\frac{cos\ 4x}{4}\ -\ \frac{cos\ 2x}{2}\ +\ c}\]
\[11.\ Evaluate:\ (i)\ \int (tan\ x\ +\ cot\ x)^2\ dx\ \hspace{2cm}\ (ii)\ \int sin^3\ x\ dx\ \hspace{8cm}\]

Soln

\[(i)\ \int (tan\ x\ +\ cot\ x)^2 \ dx\ =\ \int(tan^2\ x\ +\ cot^2\ x\ +\ 2\ tan\ x\ cot\ x)\ dx\]
\[= \int\ tan^2\ x\ dx\ +\ \int cot^2\ x\ dx\ +\ \int 2\ dx\]
\[= \int\ (sec^2\ x\ -\ 1)dx\ +\ \int (cosec^2\ x\ -\ 1)dx\ +\ 2\int dx\]
\[=\ \ tan\ x\ -\ x\ +\ cot\ x\ -\ x\ +\ 2\ x\ +\ c\]
\[=\ \ tan\ x\ +\ cot\ x\ +\ c\]
\[\therefore\ \boxed{\int(tan\ x\ +\ cot\ x)^2\ dx\ =\ tan\ x\ +\ cot\ x\ +\ c}\]
\[(ii)\ \int sin^3\ x dx\ \hspace{10cm}\]
\[W.\ K.\ T\ sin\ 3\ x\ =\ 3\ sin\ x\ -\ 4\ sin^3\ x\]
\[sin^3\ x\ =\ \frac{1}{4}[3\ sin\ x\ -\ sin\ 3\ x]\]
\[\int sin^3\ x\ dx\ =\ \int \frac{1}{4}[3\ sin\ x\ -\ sin\ 3\ x]\ dx\]
\[=\frac{1}{4}[3\int sin\ x\ dx – \int sin\ 3x\ dx]\]
\[=\frac{1}{4}[ -\ 3\ cos\ x\ +\ \frac{cos\ 3x}{3}]\ + c\]
\[\therefore\ \boxed{\int sin^3 x \ dx\ =\ \frac{1}{4}[-\ 3\ cos\ x\ +\ \frac{cos\ 3x}{3}] + c}\]