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Question Number 154195 by rexford last updated on 15/Sep/21

∫((x+sinx)/(1+cosx))dx please,help me

$$\int\frac{{x}+{sinx}}{\mathrm{1}+{cosx}}{dx}\:\:\:\: \\ $$$${please},{help}\:{me} \\ $$

Answered by qaz last updated on 15/Sep/21

∵ (((sin x)/(1+cos x)))′=(1/(1+cos x)) ∴∫((x+sin x)/(1+cos x))dx =∫(x/(1+cos x))dx+∫((sin x)/(1+cos x))dx =∫xd(((sin x)/(1+cos x)))+∫((sin x)/(1+cos x))dx =((xsin x)/(1+cos x))+C

$$\because\:\left(\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}\right)'=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cos}\:\mathrm{x}} \\ $$$$\therefore\int\frac{\mathrm{x}+\mathrm{sin}\:\mathrm{x}}{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}\mathrm{dx} \\ $$$$=\int\frac{\mathrm{x}}{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}\mathrm{dx}+\int\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}\mathrm{dx} \\ $$$$=\int\mathrm{xd}\left(\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}\right)+\int\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}\mathrm{dx} \\ $$$$=\frac{\mathrm{xsin}\:\mathrm{x}}{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}+\mathrm{C} \\ $$

Commented by qaz last updated on 15/Sep/21

∫((x−sin x)/(1+cos x))dx......∫((x−sin x)/(1−cos x))dx......∫((x+cos x)/(1+sin x))dx.....∫((x−cos x)/(1+sin x))dx ∫((x+cos x)/(1−sin x))dx......∫((x−cos x)/(1−sin x))dx.......∫((x+sin x)/(1−cos x))dx The same ....

$$\int\frac{\mathrm{x}−\mathrm{sin}\:\mathrm{x}}{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}\mathrm{dx}......\int\frac{\mathrm{x}−\mathrm{sin}\:\mathrm{x}}{\mathrm{1}−\mathrm{cos}\:\mathrm{x}}\mathrm{dx}......\int\frac{\mathrm{x}+\mathrm{cos}\:\mathrm{x}}{\mathrm{1}+\mathrm{sin}\:\mathrm{x}}\mathrm{dx}.....\int\frac{\mathrm{x}−\mathrm{cos}\:\mathrm{x}}{\mathrm{1}+\mathrm{sin}\:\mathrm{x}}\mathrm{dx} \\ $$$$\int\frac{\mathrm{x}+\mathrm{cos}\:\mathrm{x}}{\mathrm{1}−\mathrm{sin}\:\mathrm{x}}\mathrm{dx}......\int\frac{\mathrm{x}−\mathrm{cos}\:\mathrm{x}}{\mathrm{1}−\mathrm{sin}\:\mathrm{x}}\mathrm{dx}.......\int\frac{\mathrm{x}+\mathrm{sin}\:\mathrm{x}}{\mathrm{1}−\mathrm{cos}\:\mathrm{x}}\mathrm{dx} \\ $$$$\mathrm{The}\:\mathrm{same}\:.... \\ $$

Answered by puissant last updated on 15/Sep/21

Ω=∫((x+sinx)/(1+cosx))dx =∫(x/(1+cosx))dx+∫((sinx)/(1+cosx))dx =∫xsec^2 ((x/2))dx−ln∣cosx∣+C =xtan((x/2))−∫tan((x/2))dx−ln∣cosx∣+C =xtan((x/2))+2ln∣cos((x/2))∣−ln∣cosx∣+C ∴∵ Ω=xtan((x/2))+2ln∣cos((x/2))∣−ln∣cosx∣+C..

$$\Omega=\int\frac{{x}+{sinx}}{\mathrm{1}+{cosx}}{dx} \\ $$$$=\int\frac{{x}}{\mathrm{1}+{cosx}}{dx}+\int\frac{{sinx}}{\mathrm{1}+{cosx}}{dx} \\ $$$$=\int{xsec}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right){dx}−{ln}\mid{cosx}\mid+{C} \\ $$$$={xtan}\left(\frac{{x}}{\mathrm{2}}\right)−\int{tan}\left(\frac{{x}}{\mathrm{2}}\right){dx}−{ln}\mid{cosx}\mid+{C} \\ $$$$={xtan}\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{2}{ln}\mid{cos}\left(\frac{{x}}{\mathrm{2}}\right)\mid−{ln}\mid{cosx}\mid+{C} \\ $$$$ \\ $$$$\therefore\because\:\Omega={xtan}\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{2}{ln}\mid{cos}\left(\frac{{x}}{\mathrm{2}}\right)\mid−{ln}\mid{cosx}\mid+{C}.. \\ $$

Answered by maged last updated on 15/Sep/21

I=∫(x/(1+cosx)).((1−cosx)/(1−cosx))dx+∫((sinx)/(1+cosx)).((1−cosx)/(1−cosx))dx =∫((x(1−cosx))/(sin^2 x))dx+∫(1/(sinx))dx I=∫xcosec^2 xdx−∫xcotxcosecxdx+∫cosecxdx u=x→du=dx dv=cosec^2 xdx→v=−cotx −− u=x→du=dx dv=−cotxcosecxdx→v=cosecx −− I=−xcotx+xcosecx+∫cotxdx−∫cosecxdx+∫cosecxdx =xcosecx−xcotx−ln(sinx)+C

$${I}=\int\frac{{x}}{\mathrm{1}+\mathrm{cos}{x}}.\frac{\mathrm{1}−\mathrm{cos}{x}}{\mathrm{1}−\mathrm{cos}{x}}{dx}+\int\frac{\mathrm{sin}{x}}{\mathrm{1}+\mathrm{cos}{x}}.\frac{\mathrm{1}−\mathrm{cos}{x}}{\mathrm{1}−\mathrm{cos}{x}}{dx} \\ $$$$=\int\frac{{x}\left(\mathrm{1}−\mathrm{cos}{x}\right)}{\mathrm{sin}^{\mathrm{2}} {x}}{dx}+\int\frac{\mathrm{1}}{\mathrm{sin}{x}}{dx} \\ $$$${I}=\int{x}\mathrm{cosec}^{\mathrm{2}} {xdx}−\int{x}\mathrm{cot}{x}\mathrm{cosec}{xdx}+\int\mathrm{cosec}{xdx} \\ $$$${u}={x}\rightarrow{du}={dx} \\ $$$${dv}=\mathrm{cosec}^{\mathrm{2}} {xdx}\rightarrow{v}=−\mathrm{cot}{x}\: \\ $$$$−− \\ $$$${u}={x}\rightarrow{du}={dx} \\ $$$${dv}=−\mathrm{cot}{x}\mathrm{cosec}{xdx}\rightarrow{v}=\mathrm{cosec}{x} \\ $$$$−− \\ $$$${I}=−{x}\mathrm{cot}{x}+{x}\mathrm{cosec}{x}+\int\mathrm{cot}{xdx}−\int\mathrm{cosec}{xdx}+\int\mathrm{cosec}{xdx} \\ $$$$={x}\mathrm{cosec}{x}−{x}\mathrm{cot}{x}−\mathrm{ln}\left(\mathrm{sin}{x}\right)+{C} \\ $$

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