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Question Number 93937 by seedhamaieng@gmail.com last updated on 16/May/20

∫(1/(√(tan x)))dx=?

$$\int\frac{\mathrm{1}}{\sqrt{\mathrm{tan}\:{x}}}{dx}=? \\ $$

Commented by i jagooll last updated on 16/May/20

∫ ((sec^2 x)/(sec^2 x (√(tan x)))) dx = ∫ (du/((1+u^2 )(√u))) , [ u = tan x ]

$$\int\:\frac{\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}\:\sqrt{\mathrm{tan}\:\mathrm{x}}}\:\mathrm{dx}\:= \\ $$$$\int\:\frac{\mathrm{du}}{\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)\sqrt{\mathrm{u}}}\:,\:\:\left[\:\mathrm{u}\:=\:\mathrm{tan}\:\mathrm{x}\:\right] \\ $$$$ \\ $$

Commented by seedhamaieng@gmail.com last updated on 16/May/20

thanks

Commented by prakash jain last updated on 16/May/20

(√(tan x))=u tan x=u^2 sec^2 xdx=2udu (1+tan^2 x)dx=2udu (1+u^4 )dx=2udu dx=((2udu)/((1+u^4 ))) substituting value of x and dx in integral ∫(1/u)×((2u)/((1+u^4 )))du =∫(2/(1+u^4 ))du =∫((u^2 +1+1−u^2 )/(1+u^4 ))du =∫((u^2 +1)/(1+u^4 ))du−∫((u^2 −1)/(1+u^4 ))du =∫((1+(1/u^2 ))/(u^2 +(1/u^2 )−2+2))du−∫((1−(1/u^2 ))/(u^2 +(1/u^2 )+2−2))du =∫((1+(1/u^2 ))/((u−(1/u))^2 +2))du−∫((1−(1/u^2 ))/((u+(1/u))^2 −2))du in first integrl v=u−(1/u) in second integrl v=u+(1/u) =∫(dv/(v^2 +2))+∫(dv/(v^2 −2)) Now the integral can be solved using formulas for (1/(x^2 +a^2 )),(1/(x^2 −a^2 )) subtitute v with u and then x to goet final answer

$$\sqrt{\mathrm{tan}\:{x}}={u} \\ $$$$\mathrm{tan}\:{x}={u}^{\mathrm{2}} \\ $$$$\mathrm{sec}^{\mathrm{2}} {xdx}=\mathrm{2}{udu} \\ $$$$\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {x}\right){dx}=\mathrm{2}{udu} \\ $$$$\left(\mathrm{1}+{u}^{\mathrm{4}} \right){dx}=\mathrm{2}{udu} \\ $$$${dx}=\frac{\mathrm{2}{udu}}{\left(\mathrm{1}+{u}^{\mathrm{4}} \right)}\: \\ $$$$\mathrm{substituting}\:\mathrm{value}\:\mathrm{of}\:{x}\:\mathrm{and}\:\mathrm{d}{x}\:\mathrm{in}\:\mathrm{integral} \\ $$$$\int\frac{\mathrm{1}}{{u}}×\frac{\mathrm{2}{u}}{\left(\mathrm{1}+{u}^{\mathrm{4}} \right)}{du} \\ $$$$=\int\frac{\mathrm{2}}{\mathrm{1}+{u}^{\mathrm{4}} }{du} \\ $$$$=\int\frac{{u}^{\mathrm{2}} +\mathrm{1}+\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{4}} }{du} \\ $$$$=\int\frac{{u}^{\mathrm{2}} +\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{4}} }{du}−\int\frac{{u}^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{4}} }{du} \\ $$$$=\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}{{u}^{\mathrm{2}} +\frac{\mathrm{1}}{{u}^{\mathrm{2}} }−\mathrm{2}+\mathrm{2}}{du}−\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}{{u}^{\mathrm{2}} +\frac{\mathrm{1}}{{u}^{\mathrm{2}} }+\mathrm{2}−\mathrm{2}}{du} \\ $$$$=\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}{\left({u}−\frac{\mathrm{1}}{{u}}\right)^{\mathrm{2}} +\mathrm{2}}{du}−\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}{\left({u}+\frac{\mathrm{1}}{{u}}\right)^{\mathrm{2}} −\mathrm{2}}{du} \\ $$$${in}\:{first}\:{integrl}\:{v}={u}−\frac{\mathrm{1}}{{u}} \\ $$$${in}\:{second}\:{integrl}\:{v}={u}+\frac{\mathrm{1}}{{u}} \\ $$$$=\int\frac{{dv}}{{v}^{\mathrm{2}} +\mathrm{2}}+\int\frac{{dv}}{{v}^{\mathrm{2}} −\mathrm{2}} \\ $$$$\mathrm{Now}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved} \\ $$$$\mathrm{using}\:\mathrm{formulas}\:\mathrm{for} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} },\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }\: \\ $$$$\mathrm{subtitute}\:{v}\:\mathrm{with}\:{u}\:\mathrm{and}\:\mathrm{then}\:{x}\:\mathrm{to} \\ $$$$\mathrm{g}{o}\mathrm{et}\:\mathrm{final}\:\mathrm{answer} \\ $$

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 =∫ (dx/(√(tanx))) changement (√(tanx))=t give tanx =t^2 ⇒x =arctan(t^2 ) I =∫ ((2t)/((1+t^4 )t))dt =∫ ((2dt)/(1+t^4 )) =∫ ((2/t^2 )/((1/t^2 ) +t^2 ))dt =∫ ((1+(1/t^2 )+1−(1/t^2 ))/(t^2 +(1/t^2 )))dt =∫ ((1+(1/t^2 ))/((t−(1/t))^2 +2))dt +∫ ((1−(1/t^2 ))/((t+(1/t))^2 −2))dt =I_1 +I_2 I_1 =_(t−(1/t)=u) ∫ (du/(u^2 +2)) =_(u=(√2)z) ∫ (((√2)dz)/(2(1+z^2 ))) =(1/(√2)) arctan((u/(√2))) +c_1 =(1/(√2))arctan((1/(√2))(t−(1/t)))+c_1 =(1/(√2)) arctan((1/(√2))((√(tanx))−(1/(√(tanx))))) +c_1 I_2 =_(t+(1/t)=z) ∫ (dz/(z^2 −2)) =∫ (dz/((z−(√2))(z+(√2)))) =(1/(2(√2)))∫ ((1/(z−(√2)))−(1/(z+(√2))))dz =(1/(2(√2)))ln∣((z−(√2))/(z+(√2)))∣ +c_2 =(1/(2(√2)))ln∣(((√(tanx))+(1/((√(tanx)) ))−(√2))/((√(tanx))+(1/(√(tanx)))+(√2)))∣ +c_2 and I =I_1 +I_2

$${I}\:=\int\:\:\frac{{dx}}{\sqrt{{tanx}}}\:\:{changement}\:\sqrt{{tanx}}={t}\:{give}\:{tanx}\:={t}^{\mathrm{2}} \:\Rightarrow{x}\:={arctan}\left({t}^{\mathrm{2}} \right) \\ $$$${I}\:=\int\:\:\:\frac{\mathrm{2}{t}}{\left(\mathrm{1}+{t}^{\mathrm{4}} \right){t}}{dt}\:=\int\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{4}} }\:=\int\:\:\frac{\frac{\mathrm{2}}{{t}^{\mathrm{2}} }}{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\:+{t}^{\mathrm{2}} }{dt} \\ $$$$=\int\:\:\frac{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{dt}\:=\int\:\frac{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}}{dt}\:+\int\:\:\frac{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} −\mathrm{2}}{dt}\:={I}_{\mathrm{1}} +{I}_{\mathrm{2}} \\ $$$${I}_{\mathrm{1}} =_{{t}−\frac{\mathrm{1}}{{t}}={u}} \:\:\:\:\int\:\:\frac{{du}}{{u}^{\mathrm{2}} \:+\mathrm{2}}\:=_{{u}=\sqrt{\mathrm{2}}{z}} \:\:\:\int\:\:\frac{\sqrt{\mathrm{2}}{dz}}{\mathrm{2}\left(\mathrm{1}+{z}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:{arctan}\left(\frac{{u}}{\sqrt{\mathrm{2}}}\right)\:+{c}_{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{arctan}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\left({t}−\frac{\mathrm{1}}{{t}}\right)\right)+{c}_{\mathrm{1}} =\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:{arctan}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\left(\sqrt{{tanx}}−\frac{\mathrm{1}}{\sqrt{{tanx}}}\right)\right)\:+{c}_{\mathrm{1}} \\ $$$${I}_{\mathrm{2}} =_{{t}+\frac{\mathrm{1}}{{t}}={z}} \:\:\:\:\:\int\:\frac{{dz}}{{z}^{\mathrm{2}} −\mathrm{2}}\:=\int\:\:\frac{{dz}}{\left({z}−\sqrt{\mathrm{2}}\right)\left({z}+\sqrt{\mathrm{2}}\right)}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int\:\left(\frac{\mathrm{1}}{{z}−\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{{z}+\sqrt{\mathrm{2}}}\right){dz} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\mid\frac{{z}−\sqrt{\mathrm{2}}}{{z}+\sqrt{\mathrm{2}}}\mid\:+{c}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\mid\frac{\sqrt{{tanx}}+\frac{\mathrm{1}}{\sqrt{{tanx}}\:}−\sqrt{\mathrm{2}}}{\sqrt{{tanx}}+\frac{\mathrm{1}}{\sqrt{{tanx}}}+\sqrt{\mathrm{2}}}\mid\:+{c}_{\mathrm{2}} \\ $$$${and}\:{I}\:={I}_{\mathrm{1}} +{I}_{\mathrm{2}} \\ $$

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