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Question Number 73012 by vishalbhardwaj last updated on 05/Nov/19
∫(√(tanx)) dx
$$\int\sqrt{{tan}\mathrm{x}}\:{d}\mathrm{x} \\ $$
Answered by mind is power last updated on 05/Nov/19
u=(√(tan(x))) ⇒x=arctan(u^2 )⇒dx=((2u)/(1+u^4 ))du ∫((2u^2 )/(1+u^4 ))du=∫((2u^2 du)/((u^2 +(√2)u+1)(u^2 −(√2)u+1))) =(1/( (√2)))∫((−u)/(u^2 +(√2)u+1))+(u/(u^2 −(√2)u+1))du =(1/( (√2)))∫((−u−((√2)/2))/(u^2 +(√2)u+1))du+(1/( (√2)))∫((u−((√2)/2))/(u^2 −(√2)u+1))+(1/2)∫(du/((u+(1/( (√2))))^2 +(1/2))) +(1/2)∫(du/((u−(1/( (√2))))^2 +(1/2)))=(1/(2(√2)))ln(((u^2 −(√2)u+1)/(u^2 +(√2)u+1)))+(1/( (√2))) arctan((√2)u+1)+(1/( (√2)))arctan((√2)u−1)+c ∫(√(tan(x)))dx=(1/(2(√2)))(((tan(x)−(√(2tan(x)))+1)/(tan(x)+(√(2tan(x)))+1)))+(1/( (√2)))arctan(((√(2tan(x)))+1) +(1/( (√2)))arctan((√(2tan(x)))−1)
$$\mathrm{u}=\sqrt{\mathrm{tan}\left(\mathrm{x}\right)} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{arctan}\left(\mathrm{u}^{\mathrm{2}} \right)\Rightarrow\mathrm{dx}=\frac{\mathrm{2u}}{\mathrm{1}+\mathrm{u}^{\mathrm{4}} }\mathrm{du} \\ $$$$\int\frac{\mathrm{2u}^{\mathrm{2}} }{\mathrm{1}+\mathrm{u}^{\mathrm{4}} }\mathrm{du}=\int\frac{\mathrm{2u}^{\mathrm{2}} \mathrm{du}}{\left(\mathrm{u}^{\mathrm{2}} +\sqrt{\mathrm{2}}\mathrm{u}+\mathrm{1}\right)\left(\mathrm{u}^{\mathrm{2}} −\sqrt{\mathrm{2}}\mathrm{u}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{−\mathrm{u}}{\mathrm{u}^{\mathrm{2}} +\sqrt{\mathrm{2}}\mathrm{u}+\mathrm{1}}+\frac{\mathrm{u}}{\mathrm{u}^{\mathrm{2}} −\sqrt{\mathrm{2}}\mathrm{u}+\mathrm{1}}\mathrm{du} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{−\mathrm{u}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}{\mathrm{u}^{\mathrm{2}} +\sqrt{\mathrm{2}}\mathrm{u}+\mathrm{1}}\mathrm{du}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int\frac{\mathrm{u}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}{\mathrm{u}^{\mathrm{2}} −\sqrt{\mathrm{2}}\mathrm{u}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{du}}{\left(\mathrm{u}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{du}}{\left(\mathrm{u}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{ln}\left(\frac{\mathrm{u}^{\mathrm{2}} −\sqrt{\mathrm{2}}\mathrm{u}+\mathrm{1}}{\mathrm{u}^{\mathrm{2}} +\sqrt{\mathrm{2}}\mathrm{u}+\mathrm{1}}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{arctan}\left(\sqrt{\mathrm{2}}\mathrm{u}+\mathrm{1}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{arctan}\left(\sqrt{\mathrm{2}}\mathrm{u}−\mathrm{1}\right)+\mathrm{c} \\ $$$$\int\sqrt{\mathrm{tan}\left(\mathrm{x}\right)}\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left(\frac{\mathrm{tan}\left(\mathrm{x}\right)−\sqrt{\mathrm{2tan}\left(\mathrm{x}\right)}+\mathrm{1}}{\mathrm{tan}\left(\mathrm{x}\right)+\sqrt{\mathrm{2tan}\left(\mathrm{x}\right)}+\mathrm{1}}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{arctan}\left(\left(\sqrt{\mathrm{2tan}\left(\mathrm{x}\right)}+\mathrm{1}\right)\right. \\ $$$$+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{arctan}\left(\sqrt{\mathrm{2tan}\left(\mathrm{x}\right)}−\mathrm{1}\right) \\ $$$$ \\ $$
Commented by vishalbhardwaj last updated on 05/Nov/19
please give another method to solve this which I found earlier and was more easy than this method.
$$\mathrm{please}\:\mathrm{give}\:\mathrm{another}\:\mathrm{method}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{which} \\ $$$$\mathrm{I}\:\mathrm{found}\:\mathrm{earlier}\:\mathrm{and}\:\mathrm{was}\:\mathrm{more}\:\mathrm{easy}\:\mathrm{than}\:\mathrm{this}\:\mathrm{method}. \\ $$
Answered by Tanmay chaudhury last updated on 05/Nov/19
t^2 =tanx 2tdt=sec^2 xdx dx=((2tdt)/(1+t^4 )) ∫((t×2tdt)/(1+t^4 )) ∫((2dt)/(t^2 +(1/t^2 ))) concept.. ∫((1−(1/t^(2 ) )+1+(1/t^2 ))/(t^2 +(1/t^2 )))dt ∫((1−(1/t^2 ))/(t^2 +(1/t^2 )))dt+∫((1+(1/t^2 ))/(t^2 +(1/t^2 )))dt ∫((d(t+(1/t)))/((t+(1/t))^2 −2))+∫((d(t−(1/t)))/((t−(1/t))^2 +2)) =(1/(2(√2)))ln(((t+(1/t)−(√2))/(t+(1/t)+(√2))))+(1/( (√2)))tan^(−1) (((t−(1/t))/( (√2)))) now pls put t=(√(tanx))
$${t}^{\mathrm{2}} ={tanx}\:\:\:\:\:\mathrm{2}{tdt}={sec}^{\mathrm{2}} {xdx} \\ $$$${dx}=\frac{\mathrm{2}{tdt}}{\mathrm{1}+{t}^{\mathrm{4}} } \\ $$$$\int\frac{{t}×\mathrm{2}{tdt}}{\mathrm{1}+{t}^{\mathrm{4}} } \\ $$$$\int\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }} \\ $$$${concept}.. \\ $$$$\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}\:} }+\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{dt} \\ $$$$\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{dt}+\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{dt} \\ $$$$\int\frac{{d}\left({t}+\frac{\mathrm{1}}{{t}}\right)}{\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} −\mathrm{2}}+\int\frac{{d}\left({t}−\frac{\mathrm{1}}{{t}}\right)}{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}} \\ $$$$ \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\left(\frac{{t}+\frac{\mathrm{1}}{{t}}−\sqrt{\mathrm{2}}}{{t}+\frac{\mathrm{1}}{{t}}+\sqrt{\mathrm{2}}}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{{t}−\frac{\mathrm{1}}{{t}}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$${now}\:{pls}\:{put}\:{t}=\sqrt{{tanx}}\: \\ $$$$ \\ $$
Commented by Tanmay chaudhury last updated on 05/Nov/19
pls check...
$${pls}\:{check}… \\ $$
Answered by Tanmay chaudhury last updated on 05/Nov/19
method 2 p=∫(√(tanx)) +(√(cotx)) dx q=∫(√(tanx)) −(√(cotx)) dx ∫(√(tanx)) dx=((p+q)/2) now p=∫(√(tanx)) +(√(cotx)) dx =∫((sinx+cosx)/( (√(sinxcosx))))dx =(√2) ∫((d(sinx−cosx))/( (√(1−1+2sinxcosx)))) =(√2) ∫((d(sinx−cosx))/( (√(1−(sinx−cosx)^2 )))) =(√2) ×sin^(−1) (sinx−cosx) +c_1 q=∫(√(tanx)) −(√(cotx)) dx =∫((sinx−cosx)/( (√(sinxcosx))))dx=(−(√2) )∫((d(sinx+cosx))/( (√(1+2sinxcosx−1)))) =(−(√2) )∫((d(sinx+cosx))/( (√((sinx+cosx)^2 −1)))) [formula ∫(dx/( (√(x^2 −a^2 ))))) =(−(√2) )×ln[(sinx+cosx)+(√((sinx+cosx))) ]+c_2 now pls put value of p and q then find ∫(√(tanx)) dx
$${method}\:\mathrm{2} \\ $$$${p}=\int\sqrt{{tanx}}\:+\sqrt{{cotx}}\:\:{dx} \\ $$$${q}=\int\sqrt{{tanx}}\:−\sqrt{{cotx}}\:{dx} \\ $$$$\int\sqrt{{tanx}}\:{dx}=\frac{{p}+{q}}{\mathrm{2}} \\ $$$${now}\:{p}=\int\sqrt{{tanx}}\:+\sqrt{{cotx}}\:{dx} \\ $$$$=\int\frac{{sinx}+{cosx}}{\:\sqrt{{sinxcosx}}}{dx} \\ $$$$=\sqrt{\mathrm{2}}\:\int\frac{{d}\left({sinx}−{cosx}\right)}{\:\sqrt{\mathrm{1}−\mathrm{1}+\mathrm{2}{sinxcosx}}} \\ $$$$=\sqrt{\mathrm{2}}\:\int\frac{{d}\left({sinx}−{cosx}\right)}{\:\sqrt{\mathrm{1}−\left({sinx}−{cosx}\right)^{\mathrm{2}} }} \\ $$$$=\sqrt{\mathrm{2}}\:×{sin}^{−\mathrm{1}} \left({sinx}−{cosx}\right)\:\:\:+{c}_{\mathrm{1}} \\ $$$${q}=\int\sqrt{{tanx}}\:−\sqrt{{cotx}}\:{dx} \\ $$$$=\int\frac{{sinx}−{cosx}}{\:\sqrt{{sinxcosx}}}{dx}=\left(−\sqrt{\mathrm{2}}\:\right)\int\frac{{d}\left({sinx}+{cosx}\right)}{\:\sqrt{\mathrm{1}+\mathrm{2}{sinxcosx}−\mathrm{1}}} \\ $$$$=\left(−\sqrt{\mathrm{2}}\:\right)\int\frac{{d}\left({sinx}+{cosx}\right)}{\:\sqrt{\left({sinx}+{cosx}\right)^{\mathrm{2}} −\mathrm{1}}}\:\:\left[{formula}\:\:\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} \:}}\right) \\ $$$$=\left(−\sqrt{\mathrm{2}}\:\right)×{ln}\left[\left({sinx}+{cosx}\right)+\sqrt{\left({sinx}+{cosx}\right)}\:\right]+{c}_{\mathrm{2}} \\ $$$$\boldsymbol{{no}}{w}\:\:\boldsymbol{{pls}}\:\boldsymbol{{put}}\:\boldsymbol{{value}}\:\boldsymbol{{of}}\:\boldsymbol{{p}}\:\boldsymbol{{and}}\:\boldsymbol{{q}}\:\boldsymbol{{then}}\:\boldsymbol{{find}} \\ $$$$\int\sqrt{{tanx}}\:{dx} \\ $$
Commented by vishalbhardwaj last updated on 27/Nov/19
this solution is good
$$\mathrm{this}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{good} \\ $$$$ \\ $$$$ \\ $$
Answered by ajfour last updated on 27/Nov/19
I=∫(√(tan x))dx let tan x=t^2 ⇒ (1+t^4 )dx=2tdt I=∫((2t^2 )/(1+t^4 ))dt = ∫((2dt)/(t^2 +(1/t^2 ))) =∫(((1+(1/t^2 ))dt)/((t−(1/t))^2 +((√2))^2 ))+∫(((1−(1/t^2 ))dt)/((t+(1/t))^2 −((√2))^2 )) = (1/( (√2)))tan^(−1) ((((√(tan x))−(√(cot x)))/( (√2)))) +(1/(2(√2)))ln ∣(((√(tan x))+(√(cot x))−(√2))/( (√(tan x))+(√(cot x))+(√2)))∣+c .
$${I}=\int\sqrt{\mathrm{tan}\:{x}}{dx}\: \\ $$$${let}\:\:\mathrm{tan}\:{x}={t}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\left(\mathrm{1}+{t}^{\mathrm{4}} \right){dx}=\mathrm{2}{tdt} \\ $$$${I}=\int\frac{\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }{dt}\:=\:\int\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }} \\ $$$$=\int\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right){dt}}{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }+\int\frac{\left(\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right){dt}}{\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} −\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{tan}\:{x}}−\sqrt{\mathrm{cot}\:{x}}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$\:\:\:\:+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{ln}\:\mid\frac{\sqrt{\mathrm{tan}\:{x}}+\sqrt{\mathrm{cot}\:{x}}−\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{tan}\:{x}}+\sqrt{\mathrm{cot}\:{x}}+\sqrt{\mathrm{2}}}\mid+{c}\:. \\ $$$$ \\ $$

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