\begin{array}{l} \smallint \sqrt {cotx} {\rm{d}}x = \\\\ By\;Substitution\\ let \Rightarrow u = \sqrt {\cot x} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{u^2} = \cot x2u\;\\ du = - {\sec ^2}x\;dx = - \left( {1 + {{\cot }^2}x} \right)dx = - \left( {1 + {u^4}} \right)dx\\ dx = \frac{{ - 2u}}{{1 + {u^4}}}du\\ \smallint \frac{{ - 2{u^2}}}{{1 + {u^4}}}du\\ - \smallint \frac{{({u^2} + 1) + \left( {{u^2} - 1} \right)}}{{{u^4} + 1}}du = - \smallint \frac{{{u^2} + 1}}{{{u^4} + 1}}du - \smallint \frac{{{u^2} - 1}}{{{u^4} + 1}}du\\ - \smallint \frac{{1 + \frac{1}{{{u^2}}}}}{{{u^2} + \frac{1}{{{u^2}}}}}du - \smallint \frac{{1 - \frac{1}{{{u^2}}}}}{{{u^2} + \frac{1}{{{u^2}}}}}du = - \smallint \frac{{1 + \frac{1}{{{u^2}}}}}{{{{\left( {u - \frac{1}{u}} \right)}^2} + 2}}du - \smallint \frac{{1 - \frac{1}{{{u^2}}}}}{{{{\left( {u + \frac{1}{u}} \right)}^2} - 2}}du\\ by\;substitution\\ let \Rightarrow g = u - \frac{1}{u}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;dg = \left( {1 + \frac{1}{{{u^2}}}} \right)du\;\;\\ and\;let \Rightarrow v = u + \frac{1}{u}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;dv = \left( {1 - \frac{1}{{{u^2}}}} \right)du\\ - \smallint \frac{{dg}}{{{g^2} + 2}} - \smallint \frac{{dv}}{{{v^2} - 2}} = \frac{{ - 1}}{{\sqrt 2 }}{\tan ^{ - 1}}\frac{g}{{\sqrt 2 }} - \frac{1}{{2\sqrt 2 }}ln\left| {\frac{{v - \sqrt 2 }}{{v + \sqrt 2 }}} \right| + c\\ \frac{{ - 1}}{{\sqrt 2 }}{\tan ^{ - 1}}\frac{{u - \frac{1}{u}}}{{\sqrt 2 }} - \frac{1}{{2\sqrt 2 }}ln\left| {\frac{{u + \frac{1}{u}\; - \sqrt 2 }}{{u + \frac{1}{u}\; + \sqrt 2 }}} \right| + c\\ \frac{{ - 1}}{{\sqrt 2 }}{\tan ^{ - 1}}\frac{{\sqrt {\cot x} - \frac{1}{{\sqrt {\cot x} }}}}{{\sqrt 2 }} - \frac{1}{{2\sqrt 2 }}ln\left| {\frac{{\sqrt {\cot x} + \frac{1}{{\sqrt {\cot x} }}\; - \sqrt 2 }}{{\sqrt {\cot x} + \frac{1}{{\sqrt {\cot x} }}\; + \sqrt 2 }}} \right| + c \end{array}
Tags:
integration