\[Integrals\ of\ the\ form\ \int \frac{dx}{a^2 \pm x^2}, \int \frac{dx}{x^2-a^2}\ ,\ \int \frac{dx}{\sqrt{a^2 – x^2}}\]
\[\int {\sqrt{a^2 – x^2}}\ dx\ and\ \int {\sqrt{x^2 \pm a^2}}\ dx\]
List of Formulae:
\[1.\ \int \frac{dx}{a^2 + x^2} = \frac{1}{a}\ {tan}^{-1}(\frac{x}{a}) +c\ (or)\ \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\ {tan}^{-1} (\frac{x}{a}) +c\]
\[2.\ \int \frac{dx}{a^2 – x^2} = \frac{1}{2a}\ log\ (\frac{a + x}{a – x}) + c\]
\[3.\ \int \frac{dx}{x^2 – a^2} = \frac{1}{2a}\ log\ (\frac{x – a}{x + a}) + c\]
\[4.\ \int \frac{dx}{{\sqrt{a^2 – x^2}}} = {sin}^{-1}(\frac{x}{a})\ + c\]
\[5.\ \int {\sqrt{a^2 – x^2}}\ dx = \frac{x}{2}\ {\sqrt{a^2 – x^2}}\ + \frac{a^2}{2} {sin}^{-1}(\frac{x}{a})\ + c\]
\[6.\ \int {\sqrt{x^2 + a^2}}\ dx = \frac{x}{2}\ {\sqrt{x^2 + a^2}}\ + \frac{a^2}{2} {sin\ h}^{-1}(\frac{x}{a})\ + c\]
\[7.\ \int {\sqrt{x^2 – a^2}}\ dx = \frac{x}{2}\ {\sqrt{x^2 – a^2}}\ – \frac{a^2}{2} {cos\ h}^{-1}(\frac{x}{a})\ + c\]
\[\color {purple} {Example\ 1\ .}\ \color {red}{Evaluate\ :}\ \int \frac{dx}{x^2 + 4}\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[W.K.T\ \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\ {tan}^{-1} (\frac{x}{a}) +c\]
\[\int \frac{dx}{x^2 + 4} = \int \frac{dx}{x^2 + 2^2}\]
\[ = \frac{1}{2}\ {tan}^{-1} (\frac{x}{2}) +c\]
\[\boxed{\int \frac{dx}{x^2 + 4} = \frac{1}{2}\ {tan}^{-1} (\frac{x}{2}) +c}\]
\[\color {purple} {Example\ 2\ .}\ \color {red}{Evaluate\ :}\ \int \frac{dx}{49\ +\ x^2}\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[W.K.T\ \int \frac{dx}{a^2\ +\ x^2} = \frac{1}{a}\ {tan}^{-1} (\frac{x}{a}) +c\]
\[\int \frac{dx}{49\ +\ x^2}\ = \int \frac{dx}{7^2\ +\ x^2}\]
\[ = \frac{1}{7}\ {tan}^{-1} (\frac{x}{7}) +c\]
\[\boxed{\int \frac{dx}{49\ +\ x^2} = \frac{1}{7}\ {tan}^{-1} (\frac{x}{7}) +c}\]
\[\color {purple} {Example\ 3\ .}\ \color {red}{Evaluate\ :}\ \int \frac{dx}{4x^2 + 9}\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int \frac{dx}{4x^2 + 9} = \frac{1}{4}\ \int \frac{dx}{x^2 + \frac{9}{4}} = \frac{1}{4}\ \int \frac{dx}{x^2 + (\frac{3}{2})^2}\]
\[W.K.T\ \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\ {tan}^{-1} (\frac{x}{a}) +c\]
\[\int \frac{dx}{4x^2 + 9} = \frac{1}{4}\ × \frac{2}{3}\ {tan}^{-1} (\frac{x}{\frac{3}{2}}) +c\]
\[ = \frac{1}{6}\ {tan}^{-1} (\frac{2}{3}) +c\]
\[\boxed{\int \frac{dx}{4x^2 + 9}\ = \frac{1}{6}\ {tan}^{-1} (\frac{2}{3}) +c}\]
\[\color {purple} {Example\ 4\ .}\ \color {red}{Evaluate\ :}\ \int \frac{dx}{4 + 9 x^2}\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int \frac{dx}{4 + 9 x^2} = \int \frac{dx}{9 x^2+ 4} = \frac{1}{9}\ \int \frac{dx}{x^2 + \frac{4}{9}} = \frac{1}{9}\ \int \frac{dx}{x^2 + (\frac{2}{3})^2}\]
\[W.K.T\ \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\ {tan}^{-1} (\frac{x}{a}) +c\]
\[\int \frac{dx}{4 + 9 x^2} = \frac{1}{9}\ × \frac{3}{2}\ {tan}^{-1} (\frac{x}{\frac{2}{3}}) +c\]
\[ = \frac{1}{6}\ {tan}^{-1} (\frac{3x}{2}) +c\]
\[\boxed{\int \frac{dx}{4 + 9 x^2}\ = \frac{1}{6}\ {tan}^{-1} (\frac{3x}{2}) +c}\]
\[\color {purple} {Example\ 5\ .}\ \color {red}{Evaluate\ :}\ \int \frac{dx}{(3x + 2)^2 + 16}\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[Put\ u\ =\ 3x\ +\ 2\]
\[\frac{du}{dx}= \ 3\]
\[dx = \frac{1}{3}\ du\]
\[\int \frac{dx}{(3x + 2)^2 + 16} = \frac{1}{3}\ \int \frac{du}{u^2 + 4^2}\]
\[W.K.T\ \int \frac{dx}{x^2 + a^2} = \frac{1}{a}\ {tan}^{-1} (\frac{x}{a}) +c\]
\[\frac{1}{3}\ \int \frac{du}{u^2 + 4^2} = \frac{1}{3}\ × \frac{1}{4}\ {tan}^{-1} (\frac{u}{4}) +c\]
\[ = \frac{1}{12}\ {tan}^{-1} (\frac{3x + 2}{4}) +c\]
\[\boxed{\int \frac{dx}{(3x + 2)^2 + 16} = \frac{1}{12}\ {tan}^{-1} (\frac{3x + 2}{4}) +c}\]
\[\color {purple} {Example\ 6\ .}\ \color {red}{Evaluate\ :}\ \int \frac{dx}{x^2 – 36}\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\int \frac{dx}{x^2 – 36} = \int \frac{dx}{x^2 – 6^2}\]
\[W.\ K.\ T\ \int \frac{dx}{x^2 – a^2} = \frac{1}{2a}\ log\ (\frac{x – a}{x + a}) + c\]
\[\boxed{\int \frac{dx}{x^2 – 36} = \frac{1}{12}\ log\ (\frac{x – 6}{x + 6}) + c}\]
\[\color {purple} {Example\ 7\ .}\ \color {red}{Evaluate\ :}\ \int \frac{dx}{(x\ +\ 1)^2\ -\ 9}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[Put\ u\ =\ x\ +\ 1\]
\[\frac{du}{dx}= \ 1\]
\[du =\ dx\]
\[\int \frac{dx}{(x + 1)^2\ -\ 9} =\ \int \frac{du}{u^2\ -\ 3^2}\]
\[W.\ K.\ T\ \int \frac{dx}{x^2 – a^2} = \frac{1}{2a}\ log\ (\frac{x – a}{x + a}) + c\]
\[\int \frac{du}{u^2\ -\ 3^2} =\ \frac{1}{2(3)}\ log\ (\frac{u\ -\ 3}{u\ +\ 3}) +c\]
\[ =\ \frac{1}{6}\ log\ (\frac{x\ +\ 1\ -\ 3}{x\ +\ 1\ +\ 3}) +c\]
\[ = \frac{1}{6}\ log\ (\frac{x\ -\ 2}{x\ +\ 4}) +c\]
\[\boxed{\int \frac{dx}{(x + 1)^2\ -\ 9} = \frac{1}{6}\ log\ (\frac{x\ -\ 2}{x\ +\ 4}) + c}\]
\[\color {purple} {Example\ 8\ .}\ \color {red}{Evaluate\ :}\ \int \frac{dx}{(4x\ +\ 1)^2 – 36}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[Put\ u\ =\ 4x\ +\ 1\]
\[\frac{du}{dx}= \ 4\]
\[dx = \frac{1}{4}\ du\]
\[\int \frac{dx}{(4x + 1)^2\ -\ 36} = \frac{1}{4}\ \int \frac{du}{u^2\ -\ 6^2}\]
\[W.\ K.\ T\ \int \frac{dx}{x^2 – a^2} = \frac{1}{2a}\ log\ (\frac{x – a}{x + a}) + c\]
\[\frac{1}{4}\ \int \frac{du}{u^2\ -\ 6^2} = \frac{1}{4}\ × \frac{1}{2(6)}\ log\ (\frac{u\ -\ 6}{u\ +\ 6}) +c\]
\[ = \frac{1}{4}\ × \frac{1}{12}\ log\ (\frac{4x\ +\ 1\ -\ 6}{4x\ +\ 1\ +\ 6}) +c\]
\[ = \frac{1}{48}\ log\ (\frac{4x\ -\ 5}{4x\ +\ 7}) +c\]
\[\boxed{\int \frac{dx}{(4x + 1)^2\ -\ 36} = \frac{1}{48}\ log\ (\frac{4x\ -\ 5}{4x\ +\ 7}) + c}\]
\[\color {purple} {Example\ 9\ .}\ \color {red}{Evaluate\ :}\ \int \frac{dx}{9\ -\ x^2}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[\int \frac{dx}{9\ -\ x^2}\ = \ \int \frac{du}{3^2\ -\ x^2}\]
\[W.K.T\ \int \frac{dx}{a^2 – x^2} = \frac{1}{2a}\ log\ (\frac{a + x}{a – x}) + c\]
\[\int \frac{du}{3^2\ -\ x^2}\ =\ \frac{1}{2 × 3 }\ log\ (\frac{3\ +\ x}{3\ -\ x}) +c\]
\[=\frac{1}{6}\ [log\ (\frac{3\ +\ x }{3\ -\ x })] +c\]
\[\boxed{\int \frac{dx}{9\ -\ x^2} = \frac{1}{6}\ [log\ (\frac{3\ +\ x }{3\ -\ x })] + c}\]
\[\color {purple} {Example\ 10\ .}\ \color {red}{Evaluate\ :}\ \int \frac{dx}{9 – (3x – 2)^2}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
Put u = 3x – 2
\[\frac{du}{dx}= \ 3\]
\[dx = \frac{1}{3}\ du\]
\[\int \frac{dx}{9 – (3x – 2)^2}= \frac{1}{3}\ \int \frac{du}{3^2 – u^2}\]
\[W.K.T\ \int \frac{dx}{a^2 – x^2} = \frac{1}{2a}\ log\ (\frac{a + x}{a – x}) + c\]
\[\frac{1}{3}\ \int \frac{du}{3^2 – u^2} = \frac{1}{3}\ [ \frac{1}{2 × 3 }\ log\ (\frac{3 + u}{3 – u})] +c\]
\[=\frac{1}{18}\ [log\ (\frac{3 + (3x – 2)}{3 – (3x – 2)})] +c\]
\[=\frac{1}{18}\ [log\ (\frac{3 + 3x – 2}{3 – 3x + 2})] +c\]
\[=\frac{1}{18}\ [log\ (\frac{1 + 3x }{5 – 3x })] +c\]
\[\boxed{\int \frac{dx}{9 – (3x – 2)^2} = \frac{1}{18}\ [log\ (\frac{1 + 3x }{5 – 3x })] + c}\]
\[\color {purple} {Example\ 11\ .}\ \color {red}{Evaluate\ :}\ \int \frac{dx}{{\sqrt{9 – x^2}}}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[\int \frac{dx}{{\sqrt{9 – x^2}}} = \int \frac{dx}{{\sqrt{3^2 – x^2}}} \]
\[W.K.T\ \int \frac{dx}{{\sqrt{a^2 – x^2}}} = {sin}^{-1}(\frac{x}{a})\ + c\]
\[\boxed{\int \frac{dx}{{\sqrt{9 – x^2}}} = {sin}^{-1}(\frac{x}{3})\ + c}\]
\[\color {purple} {Example\ 12\ .}\ \color {red}{Evaluate\ :}\ \int \frac{dx}{{\sqrt{5 – 4x^2}}}\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[\int \frac{dx}{{\sqrt{5 – 4x^2}}} = \frac{1}{2}\ \int \frac{dx}{{\sqrt{\frac{5}{4} – x^2}}}\]
\[= \frac{1}{2}\ \int \frac{dx}{{\sqrt{(\frac{\sqrt{5}}{2})^2 – x^2}}}\]
\[W.K.T\ \int \frac{dx}{{\sqrt{a^2 – x^2}}} = {sin}^{-1}(\frac{x}{a})\ + c\]
\[\int \frac{dx}{{\sqrt{5 – 4x^2}}} = \frac{1}{2}\ {sin}^{-1}(\frac{x}{\frac{\sqrt{5}}{2}})\ + c\]
\[= \frac{1}{2}\ {sin}^{-1}(\frac{2x}{\sqrt{5}})\ + c\]
\[\boxed{\int \frac{dx}{{\sqrt{5 – 4x^2}}} = \frac{1}{2}\ {sin}^{-1}(\frac{2x}{\sqrt{5}})\ + c}\]
\[\color {purple} {Example\ 13\ .}\ \color {red}{Evaluate\ :}\ \int {\sqrt{9x^2 +16}}\ dx\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[\int {\sqrt{9x^2 +16}}\ dx =3 \int {\sqrt{x^2 +\frac{16}{9}}}\ dx = 3 \int {\sqrt{x^2 + (\frac{4}{3}}})^2\ dx\]
\[\int {\sqrt{x^2 + a^2}}\ dx = \frac{x}{2}\ {\sqrt{x^2 + a^2}}\ + \frac{a^2}{2} {sin\ h}^{-1}(\frac{x}{a})\ + c\]
\[\int {\sqrt{9x^2 +16}}\ dx = 3[\frac{x}{2}\ {\sqrt{x^2 + \frac{16}{9}}}\ + \frac{16}{18} {sin\ h}^{-1}(\frac{x}{\frac{4}{3}})]\ + c\]
\[ = 3[\frac{x}{2}\ {\sqrt{x^2 + \frac{16}{9}}}\ + \frac{8}{9} {sin\ h}^{-1}(\frac{3x}{4})]\ + c\]
\[\boxed{\int {\sqrt{9x^2 +16}}\ dx = 3[\frac{x}{2}\ {\sqrt{x^2 + \frac{16}{9}}}\ + \frac{8}{9} {sin\ h}^{-1}(\frac{3x}{4})]\ + c}\]
\[\color {purple} {Example\ 14\ .}\ \color {red}{Evaluate\ :}\ \int {\sqrt{4 -9x^2}}\ dx\ \hspace{15cm}\]
\[\color {blue}{Soln:}\ \hspace{20cm}\]
\[\int {\sqrt{4 -9x^2}}\ dx =3 \int {\sqrt{\frac{4}{9} – x^2}}\ dx = 3 \int {\sqrt{(\frac{2}{3})^2 – x^2}}\ dx\]
\[W.\ K.\ T\ \int {\sqrt{a^2 – x^2}}\ dx = \frac{x}{2}\ {\sqrt{a^2 – x^2}}\ + \frac{a^2}{2} {sin}^{-1}(\frac{x}{a})\ + c\]
\[\int {\sqrt{4 -9x^2}}\ dx = 3[\frac{x}{2}\ {\sqrt{\frac{4}{9} – x^2}}\ + \frac{4}{18} {sin}^{-1}(\frac{x}{\frac{2}{3}})]\ + c\]
\[ = 3[\frac{x}{2}\ {\sqrt{\frac{4}{9} – x^2}}\ + \frac{2}{9} {sin}^{-1}(\frac{3x}{2})]\ + c\]
\[\boxed{\int {\sqrt{4 -9x^2}}\ dx = 3[\frac{x}{2}\ {\sqrt{\frac{4}{9} – x^2}}\ + \frac{2}{9} {sin}^{-1}(\frac{3x}{2})]\ + c}\]
Exercise Problems
\[\LARGE{\color {purple} {PART- A}}\]
\[\color {purple} {1\ .}\ \color {red} {Evaluate\ :}\ \int \frac{dx}{9\ +\ x^2}\ \hspace{15cm}\]
\[\color {purple} {2\ .}\ \color {red} {Evaluate\ :}\ \int \frac{dx}{3\ +\ 2 x^2}\ \hspace{15cm}\]
\[\LARGE{\color {purple} {PART- B}}\]
\[\color {purple} {3\ .}\ \color {red} {Evaluate\ :}\ \int \frac{dx}{{\sqrt{36 – x^2}}}\ \hspace{15cm}\]
\[\color {purple} {4\ .}\ \color {red} {Evaluate\ :}\ \int \frac{dx}{4x^2\ -\ 49}\ \hspace{15cm}\]
\[\LARGE{\color {purple} {PART- C}}\]
\[\color {purple} {5\ .}\ \color {red} {Evaluate\ :}\ \hspace{2cm}\ (i)\ \int\ \frac{dx}{16\ +\ x^2}\ \hspace{2cm}\ (ii)\ \int\frac{dx}{{\sqrt{4\ -\ (x\ +\ 1)^2}}}\ \hspace{10cm}\]
\[\color {purple} {6\ .}\ \color {red} {Evaluate\ :}\ \hspace{2cm}\ (i)\ \int\ \frac{dx}{(7x + 1)^2 + 25}\ \hspace{2cm}\ (ii)\ \int \frac{dx}{{\sqrt{4\ -\ 81x^2}}}\ \hspace{10cm}\]
\[\color {purple} {7\ .}\ \color {red} {Evaluate\ :}\ \hspace{2cm}\ (i)\ \int\ \frac{dx}{64\ -\ x^2}\ \hspace{2cm}\ (ii)\ \int\frac{dx}{{\sqrt{36\ -\ (5x\ +\ 1)^2}}}\ \hspace{10cm}\]
\[\color {purple} {8\ .}\ \color {red} {Evaluate\ :}\ \int\frac{dx}{(2x + 3)^2\ +\ 49}\ \hspace{15cm}\]
\[\color {purple} {9\ .}\ \color {red} {Evaluate\ :}\ \hspace{2cm}\ (i)\ \int\ \frac{dx}{25\ -\ 9x^2}\ \hspace{2cm}\ (ii)\ \int\frac{dx}{(2x + 3)^2\ +\ 9}\ \hspace{10cm}\]
\[\color {purple} {10\ .}\ \color {red} {Evaluate\ :}\ \int\ \frac{dx}{3\ -\ 2x\ -\ x^2}\ \hspace{15cm}\]
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