STANDARD INTEGRALS (Excercise Problems with Solutions)

\[\underline{PART\ -\ A}\]
\[1.\ \color {red}{Evaluate\ :\ \int \frac{dx}{9\ +\ x^2}}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[W.K.T\ \int \frac{dx}{a^2\ +\ x^2} = \frac{1}{a}\ {tan}^{-1} (\frac{x}{a}) +c\]
\[\int \frac{dx}{9\ +\ x^2}\ = \int \frac{dx}{3^2\ +\ x^2}\]
\[ = \frac{1}{3}\ {tan}^{-1} (\frac{x}{3}) +c\]
\[\boxed{\int \frac{dx}{9\ +\ x^2} = \frac{1}{3}\ {tan}^{-1} (\frac{x}{2}) +c}\]
\[2.\ \color {red}{Evaluate\ :\ \int \frac{dx}{3\ +\ 2 x^2}}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[\int \frac{dx}{3\ +\ 2 x^2} = \int \frac{dx}{2x^2\ +\ 3} = \frac{1}{2}\ \int \frac{dx}{x^2 + \frac{3}{2}} = \frac{1}{2}\ \int \frac{dx}{x^2 + (\sqrt{\frac{3}{2}})^2}\]
\[W.K.T\ \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\ {tan}^{-1} (\frac{x}{a}) +c\]
\[\int \frac{dx}{3\ +\ 2 x^2}\ =\ \frac{1}{2}\ × \sqrt{\frac{2}{3}}\ {tan}^{-1} (\frac{x}{\sqrt{\frac{3}{2}}}) +c\]
\[ = \frac{1}{\sqrt{6}}\ {tan}^{-1} (\sqrt{\frac{2}{3}}\ x) +c\]
\[\boxed{\int \frac{dx}{3\ +\ 2 x^2}\ = \frac{1}{\sqrt{6}}\ {tan}^{-1} (\sqrt{\frac{2}{3}}\ x) +c}\]
\[\underline{PART\ -\ B}\]
\[3.\ \color {red}{Evaluate\ :\ \int \frac{dx}{{\sqrt{36 – x^2}}}}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[\int \frac{dx}{{\sqrt{36\ -\ x^2}}} = \int \frac{dx}{{\sqrt{6^2\ -\ x^2}}} \]
\[W.K.T\ \int \frac{dx}{{\sqrt{a^2 – x^2}}} = {sin}^{-1}(\frac{x}{a})\ + c\]
\[\boxed{\int \frac{dx}{{\sqrt{36\ -\ x^2}}} = {sin}^{-1}(\frac{x}{6})\ + c}\]
\[4.\ \color {red}{Evaluate\ :\ \int \frac{dx}{4x^2\ -\ 49}}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[\int \frac{dx}{4x^2\ -\ 49} = \frac{1}{4}\ \int \frac{dx}{x^2\ -\ \frac{49}{4}} = \frac{1}{4}\ \int \frac{dx}{x^2\ -\ (\frac{7}{2})^2}\]
\[W.\ K.\ T\ \int \frac{dx}{x^2 – a^2} = \frac{1}{2a}\ log\ (\frac{x – a}{x + a}) + c\]
\[\int \frac{dx}{4x^2\ -\ 49}\ =\ \frac{1}{4}[\frac{1}{2\ ×\ \frac{7}{2}}\ log\ (\frac{x\ -\ \frac{7}{2}}{x\ +\ \frac{7}{2}})] +c\]
\[=\ \frac{1}{28}\ log\ (\frac{2x\ -\ 7}{2x\ +\ 7}) + c\]
\[\boxed{\int \frac{dx}{4x^2\ -\ 49} = \frac{1}{28}\ log\ (\frac{2x\ -\ 7}{2x\ +\ 7}) + c}\]
\[\underline{PART\ -\ C}\]
\[5.\ \color{red}{Evaluate:\ \hspace{2cm}\ (i)\ \int\ \frac{dx}{16\ +\ x^2}\ \hspace{2cm}\ (ii)\ \int\frac{dx}{{\sqrt{4\ -\ (x\ +\ 1)^2}}}}\ \hspace{10cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[(i)\ \int\ \frac{dx}{16\ +\ x^2}\ dx\ \hspace{10cm}\]
\[W.K.T\ \int \frac{dx}{a^2\ +\ x^2} = \frac{1}{a}\ {tan}^{-1} (\frac{x}{a}) +c\]
\[\int \frac{dx}{16\ +\ x^2}\ = \int \frac{dx}{4^2\ +\ x^2}\]
\[ = \frac{1}{4}\ {tan}^{-1} (\frac{x}{4}) +c\]
\[\boxed{\int \frac{dx}{16\ +\ x^2} = \frac{1}{4}\ {tan}^{-1} (\frac{x}{4}) +c}\]
\[(ii)\ \int\ \frac{dx}{{\sqrt{4\ -\ (x\ +\ 1)^2}}}\ \hspace{10cm}\]
\[Put\  u\  =\ x\ +\ 1\]
\[\frac{du}{dx}= \ 1\]
\[dx = \ du\]
\[\int \frac{dx}{{\sqrt{4\ -\ (x\ +\ 1)^2}}} = \ \int \frac{du}{{\sqrt{2^2\ -\ u^2}}}\]
\[W.K.T\ \int \frac{dx}{{\sqrt{a^2 – x^2}}} = {sin}^{-1}(\frac{x}{a})\ + c\]
\[\int \frac{dx}{{\sqrt{4\ -\ (x\ +\ 1)^2}}}\ =\ {sin}^{-1}(\frac{u}{2}) \]
\[=\ {sin}^{-1}(\frac{x\ +\ 1}{2})\ + c\]
\[\boxed{\int \frac{dx}{{\sqrt{4\ -\ (x\ +\ 1)^2}}}\ = {sin}^{-1}(\frac{x\ +\ 1}{2})\ + c}\]
\[6.\ \color{red}{Evaluate:\ \hspace{2cm}\ (i)\ \int\ \frac{dx}{64\ -\ x^2}\ \hspace{2cm}\ (ii)\ \int\frac{dx}{{\sqrt{36\ -\ (5x\ +\ 1)^2}}}}\ \hspace{10cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[(i)\ \int\ \frac{dx}{64\ -\ x^2}\ \hspace{10cm}\]
\[\int \frac{dx}{64\ -\ x^2}\ = \ \int \frac{dx}{8^2\ -\ x^2}\]
\[W.K.T\ \int \frac{dx}{a^2 – x^2} = \frac{1}{2a}\ log\ (\frac{a + x}{a – x}) + c\]
\[\int \frac{dx}{64\ -\ x^2}\ = \ \frac{1}{2 × 8 }\ log\ (\frac{8\ +\ x}{8\ -\ x})\ +c\]
\[=\ \frac{1}{16}\ log\ (\frac{8\ +\ x}{8\ -\ x})\ + c\]
\[\boxed{\int \frac{dx}{64\ -\ x^2}\ = \frac{1}{16}\ log\ (\frac{8\ +\ x}{8\ -\ x})\ + c}\]
\[(ii)\ \int\ \frac{dx}{{\sqrt{36\ -\ (5x\ +\ 1)^2}}}\ \hspace{10cm}\]
\[Put\ u\ =\ 5x\ +\ 1\]
\[\frac{du}{dx}= \ 5\]
\[dx =\ \frac{1}{5} \ du\]
\[\int \frac{dx}{{\sqrt{36\ -\ (5x\ +\ 1)^2}}} = \ \frac{1}{5}\int \frac{du}{{\sqrt{6^2\ -\ u^2}}}\]
\[W.K.T\ \int \frac{dx}{{\sqrt{a^2 – x^2}}} = {sin}^{-1}(\frac{x}{a})\ + c\]
\[\int \frac{dx}{{\sqrt{36\ -\ (5x\ +\ 1)^2}}}\ =\ \frac{1}{5}\ {sin}^{-1}(\frac{u}{6}) \]
\[=\ \frac{1}{5}\ {sin}^{-1}(\frac{5x\ +\ 1}{6})\ + c\]
\[\boxed{\int \frac{dx}{{\sqrt{36\ -\ (5x\ +\ 1)^2}}}\ = \frac{1}{5}\ {sin}^{-1}(\frac{5x\ +\ 1}{6})\ + c}\]
\[7.\ \color{red}{Evaluate:\ \hspace{2cm}\ (i)\ \int\ \frac{dx}{25\ -\ 9x^2}\ \hspace{2cm}\ (ii)\ \int\frac{dx}{(2x + 3)^2\ +\ 9}}\ \hspace{10cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[(i)\ \int\ \frac{dx}{25\ -\ 9x^2}\ \hspace{10cm}\]
\[\int \frac{dx}{25\ -\ 9x^2} = \frac{1}{9}\ \int \frac{dx}{\frac{25}{9}\ -\ x^2} = \frac{1}{9}\ \int \frac{dx}{(\frac{5}{3})^2\ – x^2}\]
\[W.K.T\ \int \frac{dx}{a^2 – x^2} = \frac{1}{2a}\ log\ (\frac{a + x}{a – x}) + c\]
\[\int \frac{dx}{25\ -\ 9x^2}\ = \ \frac{1}{9}[\frac{1}{2 × \frac{5}{3}}\ log\ (\frac{\frac{5}{3}\ +\ x}{\frac{5}{3}\ -\ x})\ +c]\]
\[=\ \frac{1}{30}\ log\ (\frac{5\ +\ 3x}{5\ -\ 3x})\ + c\]
\[\boxed{\int \frac{dx}{25\ -\ 9x^2}\ = \frac{1}{30}\ log\ (\frac{5\ +\ 3x}{5\ -\ 3x})\ + c}\]
\[(ii)\ \int\ \frac{dx}{(2x + 3)^2\ +\ 9}\ \hspace{10cm}\]
\[Put\ u\ =\ 2x\ +\ 3\]
\[\frac{du}{dx}= \ 2\]
\[dx = \frac{1}{2}\ du\]
\[\int \frac{dx}{(2x + 3)^2\ +\ 9} = \frac{1}{2}\ \int \frac{du}{u^2 + 3^2}\]
\[W.K.T\ \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\ {tan}^{-1} (\frac{x}{a}) +c\]
\[\int \frac{dx}{(2x + 3)^2\ +\ 9} = \frac{1}{2}\ × \frac{1}{3}\ {tan}^{-1} (\frac{u}{3}) +c\]
\[ = \frac{1}{6}\ {tan}^{-1} (\frac{2x\ +\ 3}{4}) +c\]
\[\boxed{\int \frac{dx}{(2x + 3)^2\ +\ 9} = \frac{1}{6}\ {tan}^{-1} (\frac{2x\ +\ 3}{4}) +c}\]
\[8.\ \color{red}{Evaluate:\ \int\ \frac{dx}{3\ -\ 2x\ -\ x^2}}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[3\ -\ 2x\ -\ x^2\ =\ 3\ -\ [(x\ +\ 1)^2\ -\ 1]\]
\[=\ 4\ – (x\ +\ 1)^2\]
\[\int \frac{dx}{3\ -\ 2x\ -\ x^2} = \frac{dx}{4\ – (x\ +\ 1)^2}\]
\[Put\ u\ =\ x\ +\ 1\]
\[\frac{du}{dx}= \ 1\]
\[dx = \ du\]
\[\int \frac{dx}{3\ -\ 2x\ -\ x^2}\ =\ \int \frac{du}{2^2 – u^2}\]
\[W.K.T\ \int \frac{dx}{a^2 – x^2} = \frac{1}{2a}\ log\ (\frac{a + x}{a – x}) + c\]
\[\int \frac{dx}{3\ -\ 2x\ -\ x^2} = \ \frac{1}{2 × 2}\ log\ (\frac{2 + u}{2 – u})\ +c\]
\[=\frac{1}{4}\ log\ (\frac{2 + (x\ +\ 1)}{2\ -\ (x\ +\ 1)}) +c\]
\[=\frac{1}{4}\ log\ (\frac{2\ +\ x\ +\ 1}{2\ -\ x\ -\ 1})] +c\]
\[=\frac{1}{4}\ log\ (\frac{3\ +\ x }{1\ -\ x })\ +c\]
\[\boxed{\int \frac{dx}{3\ -\ 2x\ -\ x^2} = \frac{1}{4}\ log\ (\frac{3\ +\ x }{1\ -\ x })\ + c}\]