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Question-211708

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Question Number 211708 by Skyneless last updated on 17/Sep/24
Answered by Frix last updated on 18/Sep/24
∫(√(tan x))dx= [t=2tan^(−1) (√(tan x))] =∫((1−cos t)/(1+cos^2 t))dt ∫(1/(1+cos^2 t))dt= [u=(((√2)tan t)/2)] =((√2)/2)∫(du/(u^2 +1))=(((√2)tan^(−1) u)/2) −∫((cos t)/(1+cos^2 t))dt= [v=(((√2)sin t)/2)] =((√2)/2)∫(dv/(v^2 −1))=−(((√2)tanh^(−1) v)/2) ...which is nice somehow...
$$\int\sqrt{\mathrm{tan}\:{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{2tan}^{−\mathrm{1}} \:\sqrt{\mathrm{tan}\:{x}}\right] \\ $$$$=\int\frac{\mathrm{1}−\mathrm{cos}\:{t}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:{t}}{dt} \\ $$$$ \\ $$$$\:\:\:\:\:\int\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:{t}}{dt}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{u}=\frac{\sqrt{\mathrm{2}}\mathrm{tan}\:{t}}{\mathrm{2}}\right] \\ $$$$\:\:\:\:\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}}=\frac{\sqrt{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \:{u}}{\mathrm{2}} \\ $$$$ \\ $$$$\:\:\:\:\:−\int\frac{\mathrm{cos}\:{t}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:{t}}{dt}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{v}=\frac{\sqrt{\mathrm{2}}\mathrm{sin}\:{t}}{\mathrm{2}}\right] \\ $$$$\:\:\:\:\:\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{{dv}}{{v}^{\mathrm{2}} −\mathrm{1}}=−\frac{\sqrt{\mathrm{2}}\mathrm{tanh}^{−\mathrm{1}} \:{v}}{\mathrm{2}} \\ $$$$ \\ $$$$…\mathrm{which}\:\mathrm{is}\:\mathrm{nice}\:\mathrm{somehow}… \\ $$
Answered by BHOOPENDRA last updated on 18/Sep/24
let tanx =t^2 sec^2 x dx=2t dt (1+t^4 )dx=2tdt ⇒((2t)/((1+t^4 ))) dt=dx = ∫((2t^2 )/((1+t^4 ))) dt = ∫ ((t^2 +1)/(1+t^4 )) dt+∫((t^2 −1)/(1+t^4 )) dt = ∫((1+(1/t^2 ))/((1/t^2 )+t^2 )) dt+∫ ((1−(1/t^2 ))/((1/t^2 )+t^2 ))dt ( divided by t^2 ) =∫ ((1+(1/t^2 ))/((t−(1/t))^2 +2))dt+∫((1−(1/t^2 ))/((t+(1/t))^2 −2))dt Let =t−1/t=p ,t+1/t=q =1+(1/t^2 ) dt =dp, 1−(1/t^2 ) dt =dq =∫(1/(p^2 +2)) dp +∫(1/(q^2 −2)) dq =(1/( (√2))) tan^(−1) (p/( (√2))) +(1/(2(√2))) log(((q−(√2))/(q+(√2))))+c substituting all the values p=t−(1/t),q=t+(1/t) t=(√(tanx)) =(1/( (√2)))tan^(−1) (((tan x−1)/( (√(2tanx)))))+(1/(2(√2)))log(((tan x−(√(2tan x))+1)/(tan x+(√(2tan x))+1)))+c
$${let}\:\mathrm{tan}{x}\:={t}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\mathrm{sec}\:^{\mathrm{2}} {x}\:{dx}=\mathrm{2}{t}\:{dt} \\ $$$$\:\:\:\:\:\:\:\left(\mathrm{1}+{t}^{\mathrm{4}} \right){dx}=\mathrm{2}{tdt} \\ $$$$\Rightarrow\frac{\mathrm{2}{t}}{\left(\mathrm{1}+{t}^{\mathrm{4}} \right)}\:{dt}={dx} \\ $$$$=\:\:\int\frac{\mathrm{2}{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{4}} \right)}\:{dt} \\ $$$$=\:\int\:\:\frac{{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{4}} }\:{dt}+\int\frac{{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{4}} }\:{dt} \\ $$$$=\:\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+{t}^{\mathrm{2}} }\:{dt}+\int\:\frac{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+{t}^{\mathrm{2}} }{dt}\:\left(\:{divided}\:{by}\:{t}^{\mathrm{2}} \right) \\ $$$$=\int\:\frac{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}}{dt}+\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} −\mathrm{2}}{dt} \\ $$$${Let}\:={t}−\mathrm{1}/{t}={p}\:,{t}+\mathrm{1}/{t}={q} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\:{dt}\:={dp},\:\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\:{dt}\:={dq} \\ $$$$=\int\frac{\mathrm{1}}{{p}^{\mathrm{2}} +\mathrm{2}}\:{dp}\:+\int\frac{\mathrm{1}}{{q}^{\mathrm{2}} −\mathrm{2}}\:{dq} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{tan}^{−\mathrm{1}} \frac{{p}}{\:\sqrt{\mathrm{2}}}\:+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:{log}\left(\frac{{q}−\sqrt{\mathrm{2}}}{{q}+\sqrt{\mathrm{2}}}\right)+{c} \\ $$$${substituting}\:{all}\:{the}\:{values} \\ $$$${p}={t}−\frac{\mathrm{1}}{{t}},{q}={t}+\frac{\mathrm{1}}{{t}} \\ $$$${t}=\sqrt{{tanx}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{tan}\:{x}−\mathrm{1}}{\:\sqrt{\mathrm{2}{tanx}}}\right)+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{log}\left(\frac{\mathrm{tan}\:{x}−\sqrt{\mathrm{2tan}\:{x}}+\mathrm{1}}{\mathrm{tan}\:{x}+\sqrt{\mathrm{2tan}\:{x}}+\mathrm{1}}\right)+{c} \\ $$

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