\begin{array}{l} \smallint \sqrt {tanx} {\rm{d}}x \;\\\\ BySubstitution\\ z = \sqrt {tanx} \\ {z^2} = tanx\\ 2zdz = se{c^2}xdx\\ = (ta{n^2}x + 1)dx = ({z^4} + 1)dx\\ \;\;\;\;dx = \frac{{2z\;}}{{{z^4} + 1}}dz\\ \smallint \frac{{2{z^2}}}{{{z^4} + 1}}dz\\ \smallint \frac{{({z^2} + 1) + \left( {{z^2} - 1} \right)}}{{{z^4} + 1}}dz = \smallint \frac{{{z^2} + 1}}{{{z^4} + 1}}dz + \smallint \frac{{{z^2} - 1}}{{{z^4} + 1}}dz\\ \smallint \frac{{1 + \frac{1}{{{z^2}}}}}{{{z^2} + \frac{1}{{{z^2}}}}}dz + \smallint \frac{{1 - \frac{1}{{{z^2}}}}}{{{z^2} + \frac{1}{{{z^2}}}}}dz = \smallint \frac{{1 + \frac{1}{{{z^2}}}}}{{{{\left( {z - \frac{1}{z}} \right)}^2} + 2}}dz + \smallint \frac{{1 - \frac{1}{{{z^2}}}}}{{{{\left( {z + \frac{1}{z}} \right)}^2} - 2}}dz\\ by\;substitution\\ let \Rightarrow u = z - \frac{1}{z}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;du = \left( {1 + \frac{1}{{{z^2}}}} \right)dz\;\\ \;and\;let \Rightarrow v = z + \frac{1}{z}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;dv = \left( {1 - \frac{1}{{{z^2}}}} \right)dz\\ \smallint \frac{{du}}{{{u^2} + 2}} + \smallint \frac{{dv}}{{{v^2} - 2}} = \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\frac{u}{{\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{{z - \frac{1}{z}}}{{\sqrt 2 }} + \frac{1}{{2\sqrt 2 }}ln\left| {\frac{{z + \frac{1}{z} - \sqrt 2 }}{{z + \frac{1}{z} + \sqrt 2 }}} \right| + c\\ \frac{1}{{\sqrt 2 }}{\tan ^{ - 1}}\frac{{\sqrt {\tan x\;} - \frac{1}{{\sqrt {\tan x\;} }}}}{{\sqrt 2 }} + \frac{1}{{2\sqrt 2 }}ln\left| {\frac{{\sqrt {\tan x\;} + \frac{1}{{\sqrt {\tan x\;} }} - \sqrt 2 }}{{\sqrt {\tan x\;} + \frac{1}{{\sqrt {\tan x\;} }} + \sqrt 2 }}} \right| + c \end{array}
Tags:
integration