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Question Number 40717 by ajeetyadav4370 last updated on 26/Jul/18

∫(√(tanx/sinx.cosxdx))

$$\int\sqrt{{tanx}/{sinx}.{cosxdx}} \\ $$

Commented by math khazana by abdo last updated on 26/Jul/18

let I = ∫ (√((tanx)/(sinx cosx)))dx I = ∫ (√((sinx)/(sinx cos^2 x)))dx =∫ (dx/(cosx)) changement tan((x/2)) =tgive I = ∫ (1/((1−t^2 )/(1+t^2 ))) ((2dt)/(1+t^2 )) = ∫ ((2dt)/(1−t^2 )) = ∫ { (1/(1+t)) +(1/(1−t))}dt = ln∣((1+t)/(1−t))∣ +c =ln∣ ((1+tan((x/2)))/(1−tan((x/2))))∣+c =ln∣tan((x/2) +(π/4))∣ +c .

$${let}\:{I}\:\:=\:\int\:\:\sqrt{\frac{{tanx}}{{sinx}\:{cosx}}}{dx} \\ $$$${I}\:=\:\int\:\:\:\sqrt{\frac{{sinx}}{{sinx}\:{cos}^{\mathrm{2}} {x}}}{dx} \\ $$$$=\int\:\:\:\:\frac{{dx}}{{cosx}}\:\:\:{changement}\:\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)\:={tgive} \\ $$$${I}\:=\:\int\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\:\int\:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$$=\:\int\:\:\left\{\:\frac{\mathrm{1}}{\mathrm{1}+{t}}\:+\frac{\mathrm{1}}{\mathrm{1}−{t}}\right\}{dt}\:=\:{ln}\mid\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\mid\:+{c} \\ $$$$={ln}\mid\:\frac{\mathrm{1}+{tan}\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{1}−{tan}\left(\frac{{x}}{\mathrm{2}}\right)}\mid+{c}\:={ln}\mid{tan}\left(\frac{{x}}{\mathrm{2}}\:+\frac{\pi}{\mathrm{4}}\right)\mid\:+{c}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 26/Jul/18

∫(√((tanx)/(sinxcosx))) dx ∫secx dx ∫(dx/(cosx)) ∫((1+tan^2 (x/2))/(1−tan^2 (x/2)))dx let t=tan(x/2) dt=sec^2 (x/2).(1/2).dx ∫((1+t^2 )/(1−t^2 ))×((2dt)/(1+t^2 )) 2∫(dt/(1−t^2 )) ∫((1+t+1−t)/((1+t)(1−t)))dt ∫(dt/(1−t))+∫(dt/(1+t)) ln(1+t)−ln(1−t)+c ln(((1+t)/(1−t)))+c ln(((1+tan(x/2))/(1−tan(x/2))))+c ln(((1+2tan(x/2)+tan^2 (x/2))/(1−tan^2 (x/2))))+c ln(((1+((2tan(x/2))/(1+tan^2 (x/2))))/((1−tan^2 (x/2))/(1+tan^2 (x/2)))))=ln(((1+sinx)/(cosx)))=ln(secx+tanx)

$$\int\sqrt{\frac{{tanx}}{{sinxcosx}}}\:\:{dx} \\ $$$$\int{secx}\:{dx} \\ $$$$\int\frac{{dx}}{{cosx}} \\ $$$$\int\frac{\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{1}−{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{dx}\:\:\:{let}\:{t}={tan}\frac{{x}}{\mathrm{2}}\:\:{dt}={sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{2}}.{dx} \\ $$$$\int\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} }×\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\mathrm{2}\int\frac{{dt}}{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$$\int\frac{\mathrm{1}+{t}+\mathrm{1}−{t}}{\left(\mathrm{1}+{t}\right)\left(\mathrm{1}−{t}\right)}{dt} \\ $$$$\int\frac{{dt}}{\mathrm{1}−{t}}+\int\frac{{dt}}{\mathrm{1}+{t}} \\ $$$${ln}\left(\mathrm{1}+{t}\right)−{ln}\left(\mathrm{1}−{t}\right)+{c} \\ $$$${ln}\left(\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\right)+{c} \\ $$$${ln}\left(\frac{\mathrm{1}+{tan}\frac{{x}}{\mathrm{2}}}{\mathrm{1}−{tan}\frac{{x}}{\mathrm{2}}}\right)+{c} \\ $$$${ln}\left(\frac{\mathrm{1}+\mathrm{2}{tan}\frac{{x}}{\mathrm{2}}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{1}−{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}\right)+{c} \\ $$$${ln}\left(\frac{\mathrm{1}+\frac{\mathrm{2}{tan}\frac{{x}}{\mathrm{2}}}{\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}}{\frac{\mathrm{1}−{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}}\right)={ln}\left(\frac{\mathrm{1}+{sinx}}{{cosx}}\right)={ln}\left({secx}+{tanx}\right) \\ $$$$ \\ $$

Commented by $@ty@m last updated on 26/Jul/18

∫sec xdx ∫sec x.(((sec x+tan x))/((sec x+tan x)))dx =ln (sec x+tan x)+C

$$\int\mathrm{sec}\:{xdx} \\ $$$$\int\mathrm{sec}\:{x}.\frac{\left(\mathrm{sec}\:{x}+\mathrm{tan}\:{x}\right)}{\left(\mathrm{sec}\:{x}+\mathrm{tan}\:{x}\right)}{dx} \\ $$$$=\mathrm{ln}\:\left(\mathrm{sec}\:{x}+\mathrm{tan}\:{x}\right)+{C} \\ $$

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