How to Integrate sqrt(tan x)
Published On :2021-01-27 21:55:00
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}{{2sqrt 2 }}lnleft| {frac{{v - sqrt 2 }}{{v + sqrt 2 }}} right| + c\ frac{1}{{sqrt 2 }}{tan ^{ - 1}}frac{{z - frac{1}{z}}}{{sqrt 2 }} + frac{1}{{2sqrt 2 }}lnleft| {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}{{2sqrt 2 }}lnleft| {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}