Question Number 93937 by seedhamaieng@gmail.com last updated on 16/May/20
$$\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{tan}\:{x}}}{dx}=? \\ $$
Commented by i jagooll last updated on 16/May/20
$$\int\:\frac{\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}\:\sqrt{\mathrm{tan}\:\mathrm{x}}}\:\mathrm{dx}\:= \\ $$$$\int\:\frac{\mathrm{du}}{\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)\sqrt{\mathrm{u}}}\:,\:\:\left[\:\mathrm{u}\:=\:\mathrm{tan}\:\mathrm{x}\:\right] \\ $$$$ \\ $$
Commented by seedhamaieng@gmail.com last updated on 16/May/20
thanks
Commented by prakash jain last updated on 16/May/20
$$\sqrt{\mathrm{tan}\:{x}}={u} \\ $$$$\mathrm{tan}\:{x}={u}^{\mathrm{2}} \\ $$$$\mathrm{sec}^{\mathrm{2}} {xdx}=\mathrm{2}{udu} \\ $$$$\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {x}\right){dx}=\mathrm{2}{udu} \\ $$$$\left(\mathrm{1}+{u}^{\mathrm{4}} \right){dx}=\mathrm{2}{udu} \\ $$$${dx}=\frac{\mathrm{2}{udu}}{\left(\mathrm{1}+{u}^{\mathrm{4}} \right)}\: \\ $$$$\mathrm{substituting}\:\mathrm{value}\:\mathrm{of}\:{x}\:\mathrm{and}\:\mathrm{d}{x}\:\mathrm{in}\:\mathrm{integral} \\ $$$$\int\frac{\mathrm{1}}{{u}}×\frac{\mathrm{2}{u}}{\left(\mathrm{1}+{u}^{\mathrm{4}} \right)}{du} \\ $$$$=\int\frac{\mathrm{2}}{\mathrm{1}+{u}^{\mathrm{4}} }{du} \\ $$$$=\int\frac{{u}^{\mathrm{2}} +\mathrm{1}+\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{4}} }{du} \\ $$$$=\int\frac{{u}^{\mathrm{2}} +\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{4}} }{du}−\int\frac{{u}^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{4}} }{du} \\ $$$$=\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}{{u}^{\mathrm{2}} +\frac{\mathrm{1}}{{u}^{\mathrm{2}} }−\mathrm{2}+\mathrm{2}}{du}−\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}{{u}^{\mathrm{2}} +\frac{\mathrm{1}}{{u}^{\mathrm{2}} }+\mathrm{2}−\mathrm{2}}{du} \\ $$$$=\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}{\left({u}−\frac{\mathrm{1}}{{u}}\right)^{\mathrm{2}} +\mathrm{2}}{du}−\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}{\left({u}+\frac{\mathrm{1}}{{u}}\right)^{\mathrm{2}} −\mathrm{2}}{du} \\ $$$${in}\:{first}\:{integrl}\:{v}={u}−\frac{\mathrm{1}}{{u}} \\ $$$${in}\:{second}\:{integrl}\:{v}={u}+\frac{\mathrm{1}}{{u}} \\ $$$$=\int\frac{{dv}}{{v}^{\mathrm{2}} +\mathrm{2}}+\int\frac{{dv}}{{v}^{\mathrm{2}} −\mathrm{2}} \\ $$$$\mathrm{Now}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved} \\ $$$$\mathrm{using}\:\mathrm{formulas}\:\mathrm{for} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} },\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }\: \\ $$$$\mathrm{subtitute}\:{v}\:\mathrm{with}\:{u}\:\mathrm{and}\:\mathrm{then}\:{x}\:\mathrm{to} \\ $$$$\mathrm{g}{o}\mathrm{et}\:\mathrm{final}\:\mathrm{answer} \\ $$
Commented by seedhamaieng@gmail.com last updated on 16/May/20
thanks sir
Commented by mathmax by abdo last updated on 16/May/20
$${I}\:=\int\:\:\frac{{dx}}{\:\sqrt{{tanx}}}\:\:{changement}\:\sqrt{{tanx}}={t}\:{give}\:{tanx}\:={t}^{\mathrm{2}} \:\Rightarrow{x}\:={arctan}\left({t}^{\mathrm{2}} \right) \\ $$$${I}\:=\int\:\:\:\frac{\mathrm{2}{t}}{\left(\mathrm{1}+{t}^{\mathrm{4}} \right){t}}{dt}\:=\int\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{4}} }\:=\int\:\:\frac{\frac{\mathrm{2}}{{t}^{\mathrm{2}} }}{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\:+{t}^{\mathrm{2}} }{dt} \\ $$$$=\int\:\:\frac{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{dt}\:=\int\:\frac{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}}{dt}\:+\int\:\:\frac{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} −\mathrm{2}}{dt}\:={I}_{\mathrm{1}} +{I}_{\mathrm{2}} \\ $$$${I}_{\mathrm{1}} =_{{t}−\frac{\mathrm{1}}{{t}}={u}} \:\:\:\:\int\:\:\frac{{du}}{{u}^{\mathrm{2}} \:+\mathrm{2}}\:=_{{u}=\sqrt{\mathrm{2}}{z}} \:\:\:\int\:\:\frac{\sqrt{\mathrm{2}}{dz}}{\mathrm{2}\left(\mathrm{1}+{z}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:{arctan}\left(\frac{{u}}{\:\sqrt{\mathrm{2}}}\right)\:+{c}_{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left({t}−\frac{\mathrm{1}}{{t}}\right)\right)+{c}_{\mathrm{1}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\sqrt{{tanx}}−\frac{\mathrm{1}}{\:\sqrt{{tanx}}}\right)\right)\:+{c}_{\mathrm{1}} \\ $$$${I}_{\mathrm{2}} =_{{t}+\frac{\mathrm{1}}{{t}}={z}} \:\:\:\:\:\int\:\frac{{dz}}{{z}^{\mathrm{2}} −\mathrm{2}}\:=\int\:\:\frac{{dz}}{\left({z}−\sqrt{\mathrm{2}}\right)\left({z}+\sqrt{\mathrm{2}}\right)}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int\:\left(\frac{\mathrm{1}}{{z}−\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{{z}+\sqrt{\mathrm{2}}}\right){dz} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\mid\frac{{z}−\sqrt{\mathrm{2}}}{{z}+\sqrt{\mathrm{2}}}\mid\:+{c}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\mid\frac{\sqrt{{tanx}}+\frac{\mathrm{1}}{\:\sqrt{{tanx}}\:}−\sqrt{\mathrm{2}}}{\:\sqrt{{tanx}}+\frac{\mathrm{1}}{\:\sqrt{{tanx}}}+\sqrt{\mathrm{2}}}\mid\:+{c}_{\mathrm{2}} \\ $$$${and}\:{I}\:={I}_{\mathrm{1}} +{I}_{\mathrm{2}} \\ $$