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Question Number 86615 by ar247 last updated on 29/Mar/20
∫(√(tan x ))dx
$$\int\sqrt{{tan}\:{x}\:}{dx} \\ $$
Commented by Ar Brandon last updated on 29/Mar/20
let t^2 =tan x ⇒2tdt=sec^2 xdx ⇒dx=((2t)/(1+t^4 ))dt ∫(√(tan x))dx=∫t∙((2t)/(1+t^4 ))dt=∫((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 =∫((1+(1/t^2 ))/(2+(t−(1/t))^2 ))dt+∫((1−(1/t^2 ))/((t+(1/t))^2 −2))dt =(1/2)∙((√2)/1)arctan((1/( (√2)))(t−(1/t)))−(1/2)∙((√2)/1)arctanh((1/( (√2)))(t+(1/t)))+C
$${let}\:\:{t}^{\mathrm{2}} ={tan}\:{x}\:\:\Rightarrow\mathrm{2}{tdt}={sec}^{\mathrm{2}} {xdx} \\ $$$$\Rightarrow{dx}=\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{4}} }{dt} \\ $$$$\int\sqrt{{tan}\:{x}}{dx}=\int{t}\centerdot\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{4}} }{dt}=\int\frac{\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\int\frac{{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{4}} }{dt}+\int\frac{{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{4}} }{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+{t}^{\mathrm{2}} }{dt}+\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+{t}^{\mathrm{2}} }{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\mathrm{2}+\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} }{dt}+\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} −\mathrm{2}}{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\sqrt{\mathrm{2}}}{\mathrm{1}}{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left({t}−\frac{\mathrm{1}}{{t}}\right)\right)−\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\sqrt{\mathrm{2}}}{\mathrm{1}}{arctanh}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left({t}+\frac{\mathrm{1}}{{t}}\right)\right)+{C} \\ $$$$ \\ $$$$ \\ $$
Commented by redmiiuser last updated on 30/Mar/20
canyou pls check the answer.
$${canyou}\:{pls}\:{check}\:{the} \\ $$$${answer}. \\ $$
Answered by redmiiuser last updated on 30/Mar/20
let tanx=z dz=sec^2 x.dx sec^2 x=1+z^2 ∫(((√z).dz)/((1+z^2 ))) =∫(√z).(1+z^2 )^(−1) .dz expand (1+z^2 )^(−1) by binomial and integrate. 1−z^2 +(((−1).(−2).z^4 )/(2!))+(((−1).(−2).(−3).z^6 )/(3!))+... =Σ_(n=0) ^∞ (((−1)^n .n!.z^(2n) )/(n!)) =Σ_(n=0) ^∞ (−1)^n .z^(2n) ∫(√z).Σ_(n=0) ^∞ (−1)^n .(z)^(2n) .dz =Σ_(n=0) ^∞ (((−1)^n .z^(2n+(3/2)) )/(2n+(3/2))) =Σ_(n=0) ^∞ (((−1)^n .(tan x)^(2n+(3/2)) )/(2n+(3/2)))
$${let}\:{tanx}={z} \\ $$$${dz}=\mathrm{sec}\:^{\mathrm{2}} {x}.{dx} \\ $$$${sec}^{\mathrm{2}} {x}=\mathrm{1}+{z}^{\mathrm{2}} \\ $$$$\int\frac{\sqrt{{z}}.{dz}}{\left(\mathrm{1}+{z}^{\mathrm{2}} \right)} \\ $$$$=\int\sqrt{{z}}.\left(\mathrm{1}+{z}^{\mathrm{2}} \right)^{−\mathrm{1}} .{dz} \\ $$$${expand}\:\left(\mathrm{1}+{z}^{\mathrm{2}} \right)^{−\mathrm{1}} \\ $$$${by}\:{binomial} \\ $$$${and}\:{integrate}. \\ $$$$\mathrm{1}−{z}^{\mathrm{2}} +\frac{\left(−\mathrm{1}\right).\left(−\mathrm{2}\right).{z}^{\mathrm{4}} }{\mathrm{2}!}+\frac{\left(−\mathrm{1}\right).\left(−\mathrm{2}\right).\left(−\mathrm{3}\right).{z}^{\mathrm{6}} }{\mathrm{3}!}+… \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} .{n}!.{z}^{\mathrm{2}{n}} }{{n}!} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} .{z}^{\mathrm{2}{n}} \\ $$$$\int\sqrt{{z}}.\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} .\left({z}\right)^{\mathrm{2}{n}} .{dz} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} .{z}^{\mathrm{2}{n}+\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{2}{n}+\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} .\left(\mathrm{tan}\:{x}\right)^{\mathrm{2}{n}+\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{2}{n}+\frac{\mathrm{3}}{\mathrm{2}}} \\ $$
Commented by redmiiuser last updated on 30/Mar/20
sir pls check the answer.
$${sir}\:{pls}\:{check}\:{the}\:{answer}. \\ $$

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