How to Integrate sqrt(cot x)
Published On :2021-01-27 22:07:00
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}{{2sqrt 2 }}lnleft| {frac{{v - sqrt 2 }}{{v + sqrt 2 }}} right| + c\ frac{{ - 1}}{{sqrt 2 }}{tan ^{ - 1}}frac{{u - frac{1}{u}}}{{sqrt 2 }} - frac{1}{{2sqrt 2 }}lnleft| {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}{{2sqrt 2 }}lnleft| {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}