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Question Number 44509 by arvinddayama01@gmail.com last updated on 30/Sep/18
∫(√(tan x)) dx=?
$$\int\sqrt{\boldsymbol{\mathrm{tan}}\:\boldsymbol{\mathrm{x}}}\:\:\boldsymbol{\mathrm{dx}}=? \\ $$
Commented by maxmathsup by imad last updated on 30/Sep/18
this integral is solved look at the platform ...
$${this}\:{integral}\:{is}\:{solved}\:\:{look}\:\:{at}\:{the}\:{platform}\:… \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 30/Sep/18
t^2 =tanx 2tdt=sec^2 x dx dx=((2tdt)/(sec^2 x))=((2tdt)/(1+t^4 )) ∫t×((2tdt)/(1+t^4 )) ∫((2dt)/(t^2 +(1/t^2 ))) ∫((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 p=t+(1/t) dp=1−(1/t^2 )dt q=t−(1/t) dq=1+(1/t^2 )dt ∫(dp/(p^2 −2))+∫(dq/(q^2 +2)) (1/(2(√2) ))ln(((p−(√2) )/(p+(√2) )))+(1/( (√2) ))tan^(−1) ((q/( (√2) )))+c (1/(2(√2) ))ln(((t+(1/t)−(√2) )/(t+(1/t)+(√2) )))+(1/( (√2) ))tan^(−1) (((t−(1/t))/( (√2) )))+c (1/(2(√2) ))ln((((√(tanx)) +(1/( (√(tanx)) ))−(√2))/( (√(tanx)) +(1/( (√(tanx)) ))+(√2))))+(1/( (√2) ))tan^(−1) ((((√(tanx)) −(1/( (√(tanx)) )))/( (√2))))+c
$${t}^{\mathrm{2}} ={tanx}\:\:\:\mathrm{2}{tdt}={sec}^{\mathrm{2}} {x}\:{dx} \\ $$$${dx}=\frac{\mathrm{2}{tdt}}{{sec}^{\mathrm{2}} {x}}=\frac{\mathrm{2}{tdt}}{\mathrm{1}+{t}^{\mathrm{4}} } \\ $$$$\int{t}×\frac{\mathrm{2}{tdt}}{\mathrm{1}+{t}^{\mathrm{4}} } \\ $$$$\int\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }} \\ $$$$\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{dt}\:\: \\ $$$$\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} −\mathrm{2}}{dt}+\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}}{dt} \\ $$$${p}={t}+\frac{\mathrm{1}}{{t}}\:\:{dp}=\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }{dt} \\ $$$${q}={t}−\frac{\mathrm{1}}{{t}}\:\:\:{dq}=\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }{dt} \\ $$$$\int\frac{{dp}}{{p}^{\mathrm{2}} −\mathrm{2}}+\int\frac{{dq}}{{q}^{\mathrm{2}} +\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}\:}{ln}\left(\frac{{p}−\sqrt{\mathrm{2}}\:}{{p}+\sqrt{\mathrm{2}}\:}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}{tan}^{−\mathrm{1}} \left(\frac{{q}}{\:\sqrt{\mathrm{2}}\:}\right)+{c} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}\:}{ln}\left(\frac{{t}+\frac{\mathrm{1}}{{t}}−\sqrt{\mathrm{2}}\:}{{t}+\frac{\mathrm{1}}{{t}}+\sqrt{\mathrm{2}}\:}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}{tan}^{−\mathrm{1}} \left(\frac{{t}−\frac{\mathrm{1}}{{t}}}{\:\sqrt{\mathrm{2}}\:}\right)+{c} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}\:}{ln}\left(\frac{\sqrt{{tanx}}\:+\frac{\mathrm{1}}{\:\sqrt{{tanx}}\:}−\sqrt{\mathrm{2}}}{\:\sqrt{{tanx}}\:+\frac{\mathrm{1}}{\:\sqrt{{tanx}}\:}+\sqrt{\mathrm{2}}}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}{tan}^{−\mathrm{1}} \left(\frac{\sqrt{{tanx}}\:−\frac{\mathrm{1}}{\:\sqrt{{tanx}}\:}}{\:\sqrt{\mathrm{2}}}\right)+{c} \\ $$

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