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Question Number 88339 by Rio Michael last updated on 10/Apr/20
∫( (√(tan x )) + (√(cot x)) )dx = ?
$$\:\int\left(\:\sqrt{\mathrm{tan}\:{x}\:}\:+\:\sqrt{\mathrm{cot}\:{x}}\:\right){dx}\:=\:? \\ $$
Commented by jagoll last updated on 10/Apr/20
(√(tan x)) + (1/( (√(tan x)))) = ((tan x+1)/( (√(tan x)))) (√(tan x)) = t , tan x = t^2 2sec^2 x dx = 2t dt dx = (t/(1+t^4 )) dt ] ∫ ((t^2 +1)/t)× (t/(t^4 +1)) dt = ∫ ((t^2 +1)/(t^4 +1)) dt similar ∫ ((x^2 +1)/(x^4 +1)) dx
$$\sqrt{\mathrm{tan}\:\mathrm{x}}\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{tan}\:\mathrm{x}}}\:=\:\frac{\mathrm{tan}\:\mathrm{x}+\mathrm{1}}{\:\sqrt{\mathrm{tan}\:\mathrm{x}}} \\ $$$$\sqrt{\mathrm{tan}\:\mathrm{x}}\:=\:\mathrm{t}\:,\:\mathrm{tan}\:\mathrm{x}\:=\:\mathrm{t}^{\mathrm{2}} \\ $$$$\mathrm{2sec}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{dx}\:=\:\mathrm{2t}\:\mathrm{dt} \\ $$$$\left.\mathrm{dx}\:=\:\frac{\mathrm{t}}{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\:\mathrm{dt}\:\right]\: \\ $$$$\int\:\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{t}}×\:\frac{\mathrm{t}}{\mathrm{t}^{\mathrm{4}} +\mathrm{1}}\:\mathrm{dt}\:=\:\int\:\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{t}^{\mathrm{4}} +\mathrm{1}}\:\mathrm{dt} \\ $$$$\mathrm{similar}\:\int\:\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{x}^{\mathrm{4}} +\mathrm{1}}\:\mathrm{dx} \\ $$
Commented by jagoll last updated on 10/Apr/20
∫ ((t^2 +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)) = (1/( (√2))) arc tan (((t−(1/t))/( (√2)))) + c = (1/( (√2))) arc tan ((((√(tan x)) −(1/( (√(tan x)))))/( (√2)))) +c = (1/( (√2) )) arc tan (((tan x−1)/( (√(2tan x))))) + c
$$\int\:\frac{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{t}^{\mathrm{2}} \left(\mathrm{t}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }\right)}\:\mathrm{dt}\:=\:\int\:\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}{\mathrm{t}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}\:\mathrm{dt} \\ $$$$=\:\int\:\frac{\mathrm{d}\left(\mathrm{t}−\frac{\mathrm{1}}{\mathrm{t}}\right)}{\left(\mathrm{t}−\frac{\mathrm{1}}{\mathrm{t}}\right)^{\mathrm{2}} +\mathrm{2}} \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\mathrm{t}−\frac{\mathrm{1}}{\mathrm{t}}}{\:\sqrt{\mathrm{2}}}\right)\:+\:\mathrm{c} \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\sqrt{\mathrm{tan}\:\mathrm{x}}\:−\frac{\mathrm{1}}{\:\sqrt{\mathrm{tan}\:\mathrm{x}}}}{\:\sqrt{\mathrm{2}}}\right)\:+\mathrm{c} \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\:\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\mathrm{tan}\:\mathrm{x}−\mathrm{1}}{\:\sqrt{\mathrm{2tan}\:\mathrm{x}}}\right)\:+\:\mathrm{c} \\ $$
Commented by Rio Michael last updated on 10/Apr/20
sir ,if tan x = t^2 ⇒ sec^2 x = 2t (dt/dx) ⇒ sec^2 x dx = 2t dt how did you get 2 sec^2 x dx = 2t dt??
$$\mathrm{sir}\:,\mathrm{if}\:\:\mathrm{tan}\:{x}\:=\:{t}^{\mathrm{2}} \\ $$$$\:\Rightarrow\:\mathrm{sec}^{\mathrm{2}} {x}\:=\:\mathrm{2}{t}\:\frac{{dt}}{{dx}}\:\Rightarrow\:\:\:\mathrm{sec}^{\mathrm{2}} {x}\:{dx}\:=\:\mathrm{2}{t}\:{dt} \\ $$$$\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{get}\:\:\:\mathrm{2}\:\mathrm{sec}^{\mathrm{2}} {x}\:{dx}\:=\:\mathrm{2}{t}\:{dt}?? \\ $$$$ \\ $$
Answered by TANMAY PANACEA. last updated on 10/Apr/20
method−1 ∫((√(sinx))/( (√(cosx)) ))+((√(cosx))/( (√(sinx)))) dx ∫((sinx+cosx)/((1/( (√2)))×(√(2sinxcosx))))dx (√2) ∫((d(sinx−cosx))/( (√(1−1+2sinxcosx))))dx (√2) ∫((d(sinx−cosx))/( (√(1−(sinx−cosx)^2 ))))=(√2) sin^(−1) (((sinx−cosx)/1))+c
$${method}−\mathrm{1} \\ $$$$\int\frac{\sqrt{{sinx}}}{\:\sqrt{{cosx}}\:}+\frac{\sqrt{{cosx}}}{\:\sqrt{{sinx}}}\:{dx} \\ $$$$\int\frac{{sinx}+{cosx}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}×\sqrt{\mathrm{2}{sinxcosx}}}{dx} \\ $$$$\sqrt{\mathrm{2}}\:\int\frac{{d}\left({sinx}−{cosx}\right)}{\:\sqrt{\mathrm{1}−\mathrm{1}+\mathrm{2}{sinxcosx}}}{dx} \\ $$$$\sqrt{\mathrm{2}}\:\int\frac{{d}\left({sinx}−{cosx}\right)}{\:\sqrt{\mathrm{1}−\left({sinx}−{cosx}\right)^{\mathrm{2}} }}=\sqrt{\mathrm{2}}\:{sin}^{−\mathrm{1}} \left(\frac{{sinx}−{cosx}}{\mathrm{1}}\right)+{c} \\ $$$$ \\ $$
Answered by TANMAY PANACEA. last updated on 10/Apr/20
method−2 t^2 =tanx 2tdt=sec^2 xdx dx=((2t)/(1+t^4 ))dt ∫(t+(1/t))×((2t)/(1+t^4 ))dt 2∫((t(t+(1/t)))/(t^2 (t^2 +(1/t^2 ))))dt 2∫((1+(1/t^2 ))/((t−(1/t))^2 +2))dt 2∫((d(t−(1/t)))/((t−(1/t))^2 +2))dt 2×(1/( (√2)))tan^(−1) (((t−(1/t))/( (√2))))+c=(√2) tan^(−1) ((((√(tanx)) −(√(cotx)))/( (√2))))+c
$${method}−\mathrm{2} \\ $$$${t}^{\mathrm{2}} ={tanx} \\ $$$$\mathrm{2}{tdt}={sec}^{\mathrm{2}} {xdx} \\ $$$${dx}=\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{4}} }{dt} \\ $$$$\int\left({t}+\frac{\mathrm{1}}{{t}}\right)×\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{4}} }{dt} \\ $$$$\mathrm{2}\int\frac{{t}\left({t}+\frac{\mathrm{1}}{{t}}\right)}{{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)}{dt} \\ $$$$\mathrm{2}\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}}{dt} \\ $$$$\mathrm{2}\int\frac{{d}\left({t}−\frac{\mathrm{1}}{{t}}\right)}{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}}{dt} \\ $$$$\mathrm{2}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{{t}−\frac{\mathrm{1}}{{t}}}{\:\sqrt{\mathrm{2}}}\right)+{c}=\sqrt{\mathrm{2}}\:\boldsymbol{{tan}}^{−\mathrm{1}} \left(\frac{\sqrt{\boldsymbol{{tanx}}}\:−\sqrt{\boldsymbol{{cotx}}}}{\:\sqrt{\mathrm{2}}}\right)+\boldsymbol{{c}} \\ $$
Commented by M±th+et£s last updated on 10/Apr/20
nice solution sir
$${nice}\:{solution}\:{sir} \\ $$
Answered by M±th+et£s last updated on 10/Apr/20
=∫((√(tan(x))) +(√(cot(x)))) ((tan(x)+cot(x))/(tan(x)+cot(x)))dx =∫(1/(tan(x)+cot(x))) (tan(x)(√(tan(x))) +(1/( (√(tan(x))))) + (1/( (√(cot(x)))))+cot(x)(√(cot(x))))dx =∫(1/(tan(x)+cot(x)))(((tan^2 (x)+1)/( (√(tan(x)))))+((1+cot^2 (x))/( (√(cot(x))))))dx =∫(1/(2+tan(x)−2+cot(x)))(((sec^2 (x))/( (√(tan(x)))))+((csc^2 (x))/( (√(cot(x))))))dx =∫(6/(2+((√(tan(x)))−(√(cot(x))))^2 )) (((sec^2 (x))/(2(√(tan(x)))))−((−csc^2 (x))/(2(√(cot(x))))))dx =(√2) tan^(−1) ((((√(tan(x)))−(√(cot(x))))/2))+c
$$=\int\left(\sqrt{{tan}\left({x}\right)}\:+\sqrt{{cot}\left({x}\right)}\right)\:\frac{{tan}\left({x}\right)+{cot}\left({x}\right)}{{tan}\left({x}\right)+{cot}\left({x}\right)}{dx} \\ $$$$=\int\frac{\mathrm{1}}{{tan}\left({x}\right)+{cot}\left({x}\right)}\:\left({tan}\left({x}\right)\sqrt{{tan}\left({x}\right)}\:+\frac{\mathrm{1}}{\:\sqrt{{tan}\left({x}\right)}}\:+\:\frac{\mathrm{1}}{\:\sqrt{{cot}\left({x}\right)}}+{cot}\left({x}\right)\sqrt{{cot}\left({x}\right)}\right){dx} \\ $$$$=\int\frac{\mathrm{1}}{{tan}\left({x}\right)+{cot}\left({x}\right)}\left(\frac{{tan}^{\mathrm{2}} \left({x}\right)+\mathrm{1}}{\:\sqrt{{tan}\left({x}\right)}}+\frac{\mathrm{1}+{cot}^{\mathrm{2}} \left({x}\right)}{\:\sqrt{{cot}\left({x}\right)}}\right){dx} \\ $$$$=\int\frac{\mathrm{1}}{\mathrm{2}+{tan}\left({x}\right)−\mathrm{2}+{cot}\left({x}\right)}\left(\frac{{sec}^{\mathrm{2}} \left({x}\right)}{\:\sqrt{{tan}\left({x}\right)}}+\frac{{csc}^{\mathrm{2}} \left({x}\right)}{\:\sqrt{{cot}\left({x}\right)}}\right){dx} \\ $$$$=\int\frac{\mathrm{6}}{\mathrm{2}+\left(\sqrt{{tan}\left({x}\right)}−\sqrt{{cot}\left({x}\right)}\right)^{\mathrm{2}} }\:\left(\frac{{sec}^{\mathrm{2}} \left({x}\right)}{\mathrm{2}\sqrt{{tan}\left({x}\right)}}−\frac{−{csc}^{\mathrm{2}} \left({x}\right)}{\mathrm{2}\sqrt{{cot}\left({x}\right)}}\right){dx} \\ $$$$=\sqrt{\mathrm{2}}\:{tan}^{−\mathrm{1}} \left(\frac{\sqrt{{tan}\left({x}\right)}−\sqrt{{cot}\left({x}\right)}}{\mathrm{2}}\right)+{c} \\ $$

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