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Question Number 56401 by harish 12@g last updated on 16/Mar/19
∫((x+sin x)/(1+cos x))dx
$$\int\frac{{x}+\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{cos}\:{x}}{dx} \\ $$
Commented by maxmathsup by imad last updated on 16/Mar/19
I =∫ (x/(1+cosx))dx +∫ ((sinx)/(1+cosx))dx we have ∫ ((sinx)/(1+cosx))dx =−ln∣1+cosx∣ +c_1 ∫ (x/(1+cosx))dx =_(tan((x/2))=t) ∫ ((2arctan(t))/(1+((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) =4 ∫ ((arctan(t))/(1+t^2 +1−t^2 )) dt =2 ∫ arctan(t)dt =2{tarctan(t)−∫ (t/(1+t^2 ))dt } =2{t arctan(t)−(1/2)ln(1+t^2 )} +c_2 =2tan((x/2))(x/2) −ln(1+tan^2 ((x/2)) +c_2 =x tan((x/2))−ln(1+tan^2 ((x/2))) +c_2 ⇒ I =−ln∣1+cosx∣ +x tan((x/2))−ln(1+tan^2 ((x/2))) +C .
$${I}\:=\int\:\:\frac{{x}}{\mathrm{1}+{cosx}}{dx}\:+\int\:\:\frac{{sinx}}{\mathrm{1}+{cosx}}{dx}\:\:\:\:\:\:{we}\:{have}\:\int\:\frac{{sinx}}{\mathrm{1}+{cosx}}{dx}\:=−{ln}\mid\mathrm{1}+{cosx}\mid\:+{c}_{\mathrm{1}} \\ $$$$\int\:\:\frac{{x}}{\mathrm{1}+{cosx}}{dx}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\:\:\:\:\:\int\:\:\:\frac{\mathrm{2}{arctan}\left({t}\right)}{\mathrm{1}+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\mathrm{4}\:\int\:\:\frac{{arctan}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} \:+\mathrm{1}−{t}^{\mathrm{2}} }\:{dt}\:=\mathrm{2}\:\int\:{arctan}\left({t}\right){dt}\: \\ $$$$=\mathrm{2}\left\{{tarctan}\left({t}\right)−\int\:\:\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\right\}\:=\mathrm{2}\left\{{t}\:{arctan}\left({t}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\right\}\:+{c}_{\mathrm{2}} \\ $$$$=\mathrm{2}{tan}\left(\frac{{x}}{\mathrm{2}}\right)\frac{{x}}{\mathrm{2}}\:−{ln}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\:+{c}_{\mathrm{2}} \:\:={x}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)−{ln}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)\:+{c}_{\mathrm{2}} \:\Rightarrow\right. \\ $$$${I}\:=−{ln}\mid\mathrm{1}+{cosx}\mid\:+{x}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)−{ln}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)\:+{C}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 16/Mar/19
∫((x+2sin(x/2)cos(x/2))/(2cos^2 (x/2)))dx ∫((1/2)×xsec^2 (x/2)+tan(x/2))dx ∫[x×(d/dx)(tan(x/2))+tan(x/2)×(dx/dx) ]dx ∫(d/dx)(xtan(x/2))dx ∫d(xtan(x/2)) xtan(x/2)+c
$$\int\frac{{x}+\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}}{\mathrm{2}{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{dx} \\ $$$$\int\left(\frac{\mathrm{1}}{\mathrm{2}}×{xsec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}+{tan}\frac{{x}}{\mathrm{2}}\right){dx} \\ $$$$\int\left[{x}×\frac{{d}}{{dx}}\left({tan}\frac{{x}}{\mathrm{2}}\right)+{tan}\frac{{x}}{\mathrm{2}}×\frac{{dx}}{{dx}}\:\right]{dx} \\ $$$$\int\frac{{d}}{{dx}}\left({xtan}\frac{{x}}{\mathrm{2}}\right){dx} \\ $$$$\int{d}\left({xtan}\frac{{x}}{\mathrm{2}}\right) \\ $$$${xtan}\frac{{x}}{\mathrm{2}}+{c} \\ $$
Answered by Smail last updated on 16/Mar/19
=∫((x/(1+cosx))+((sinx)/(1+cosx)))dx =∫((x/(2cos^2 (x/2)))+((2sin((x/2))cos((x/2)))/(2cos^2 (x/2))))dx =∫((x/(2cos^2 x/2))+tan(x/2))dx =∫(xd(tan(x/2))+tan(x/2)dx)=∫(vdu+udv) =∫d(vu)=vu+C with v=x and u=tan(x/2) Therefore: ∫((x+sinx)/(1+cosx))dx=xtan(x/2)+C
$$=\int\left(\frac{{x}}{\mathrm{1}+{cosx}}+\frac{{sinx}}{\mathrm{1}+{cosx}}\right){dx} \\ $$$$=\int\left(\frac{{x}}{\mathrm{2}{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}+\frac{\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right){cos}\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2}{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}\right){dx} \\ $$$$=\int\left(\frac{{x}}{\mathrm{2}{cos}^{\mathrm{2}} {x}/\mathrm{2}}+{tan}\frac{{x}}{\mathrm{2}}\right){dx} \\ $$$$=\int\left({xd}\left({tan}\frac{{x}}{\mathrm{2}}\right)+{tan}\frac{{x}}{\mathrm{2}}{dx}\right)=\int\left({vdu}+{udv}\right) \\ $$$$=\int{d}\left({vu}\right)={vu}+{C} \\ $$$${with}\:\:\:{v}={x}\:{and}\:\:{u}={tan}\frac{{x}}{\mathrm{2}} \\ $$$${Therefore}: \\ $$$$\int\frac{{x}+{sinx}}{\mathrm{1}+{cosx}}{dx}={xtan}\frac{{x}}{\mathrm{2}}+{C} \\ $$

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