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Question Number 153924 by ZiYangLee last updated on 12/Sep/21
∫ ((1+sin x)/(sin x(1+cos x))) dx =?
$$\int\:\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{sin}\:{x}\left(\mathrm{1}+\mathrm{cos}\:{x}\right)}\:{dx}\:=? \\ $$
Answered by ArielVyny last updated on 12/Sep/21
∫(((1+sinx))/(sinx(1+cosx)))dx=∫(1/(sinx(1+cosx)))+∫(1/(1+cosx))dx t=tan((x/2))→dt=(1/2)(1+t^2 )dx dx=(2/(1+t^2 ))dt ∫(1/(((2t)/(1+t^2 ))(1+((1−t^2 )/(1+t^2 )))))×(2/(1+t^2 ))dt+∫(1/(1+((1−t^2 )/(1+t^2 )))).(2/(1+t^2 ))dt ∫(1/(t(1+((1−t^2 )/(1+t^2 )))))dt+∫(1/((1+t^2 )+1−t^2 ))dt ∫(1/(t(((1+t^2 )/(1+t^2 ))+((1−t^2 )/(1+t^2 )))))dt+∫(1/2)dt ∫((1+t^2 )/(2t))dt+(1/2)t+C (1/2)ln∣t∣+(1/4)t^2 +(1/2)t+C ∫((1+sinx)/(sinx(1+cosx)))dt=ln((√(∣tan((x/2))∣)))+(1/4)(tan((x/2)))^2 +(1/2)(tan((x/2)))+C
$$\int\frac{\left(\mathrm{1}+{sinx}\right)}{{sinx}\left(\mathrm{1}+{cosx}\right)}{dx}=\int\frac{\mathrm{1}}{{sinx}\left(\mathrm{1}+{cosx}\right)}+\int\frac{\mathrm{1}}{\mathrm{1}+{cosx}}{dx} \\ $$$${t}={tan}\left(\frac{{x}}{\mathrm{2}}\right)\rightarrow{dt}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{t}^{\mathrm{2}} \right){dx} \\ $$$${dx}=\frac{\mathrm{2}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\int\frac{\mathrm{1}}{\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\left(\mathrm{1}+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)}×\frac{\mathrm{2}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}+\int\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}.\frac{\mathrm{2}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\int\frac{\mathrm{1}}{{t}\left(\mathrm{1}+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)}{dt}+\int\frac{\mathrm{1}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)+\mathrm{1}−{t}^{\mathrm{2}} }{dt} \\ $$$$\int\frac{\mathrm{1}}{{t}\left(\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)}{dt}+\int\frac{\mathrm{1}}{\mathrm{2}}{dt} \\ $$$$\int\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{2}{t}}{dt}+\frac{\mathrm{1}}{\mathrm{2}}{t}+{C} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{t}\mid+\frac{\mathrm{1}}{\mathrm{4}}{t}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{t}+{C} \\ $$$$\int\frac{\mathrm{1}+{sinx}}{{sinx}\left(\mathrm{1}+{cosx}\right)}{dt}={ln}\left(\sqrt{\mid{tan}\left(\frac{{x}}{\mathrm{2}}\right)\mid}\right)+\frac{\mathrm{1}}{\mathrm{4}}\left({tan}\left(\frac{{x}}{\mathrm{2}}\right)\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\left({tan}\left(\frac{{x}}{\mathrm{2}}\right)\right)+{C} \\ $$$$ \\ $$
Commented by Tawa11 last updated on 12/Sep/21
Weldone sir.
$$\mathrm{Weldone}\:\mathrm{sir}. \\ $$
Answered by puissant last updated on 12/Sep/21
Q=∫((1+sinx)/(sinx(1+cosx)))dx u=tan((x/2))→x=2arctan(u)→dx=((2du)/(1+u^2 )) Q=∫((1+((2u)/(1+u^2 )))/(((2u)/(1+u^2 ))(1+((1−u^2 )/(1+u^2 )))))×((2du)/(1+u^2 )) ⇒ Q=∫(((u^2 +2u+1)/(1+u^2 ))/(((2u)/(1+u^2 ))×(2/(1+u^2 ))))×((2du)/(1+u^2 )) ⇒ Q=∫((u^2 +2u+1)/(2u))du ⇒ Q=(1/4)u^2 +u+(1/2)ln∣u∣+C ∴∵ Q=(1/4)(tan((x/2)))^2 +tan((x/2))+(1/2)ln∣tan((x/2))∣+C..
$${Q}=\int\frac{\mathrm{1}+{sinx}}{{sinx}\left(\mathrm{1}+{cosx}\right)}{dx} \\ $$$${u}={tan}\left(\frac{{x}}{\mathrm{2}}\right)\rightarrow{x}=\mathrm{2}{arctan}\left({u}\right)\rightarrow{dx}=\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$${Q}=\int\frac{\mathrm{1}+\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}{\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }\left(\mathrm{1}+\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }\right)}×\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$\Rightarrow\:{Q}=\int\frac{\frac{{u}^{\mathrm{2}} +\mathrm{2}{u}+\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }}{\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }×\frac{\mathrm{2}}{\mathrm{1}+{u}^{\mathrm{2}} }}×\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$\Rightarrow\:{Q}=\int\frac{{u}^{\mathrm{2}} +\mathrm{2}{u}+\mathrm{1}}{\mathrm{2}{u}}{du} \\ $$$$\Rightarrow\:{Q}=\frac{\mathrm{1}}{\mathrm{4}}{u}^{\mathrm{2}} +{u}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{u}\mid+{C} \\ $$$$ \\ $$$$\therefore\because\:\:{Q}=\frac{\mathrm{1}}{\mathrm{4}}\left({tan}\left(\frac{{x}}{\mathrm{2}}\right)\right)^{\mathrm{2}} +{tan}\left(\frac{{x}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{tan}\left(\frac{{x}}{\mathrm{2}}\right)\mid+{C}.. \\ $$
Commented by ZiYangLee last updated on 12/Sep/21
thank you puissant...for always helping me on my integration problems...
$${thank}\:{you}\:{puissant}…{for}\:{always}\:{helping}\:{me}\:{on} \\ $$$${my}\:{integration}\:{problems}… \\ $$
Commented by puissant last updated on 12/Sep/21
Thank you. I am really happy..
$${Thank}\:{you}.\:{I}\:{am}\:{really}\:{happy}.. \\ $$

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