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x-sin-x-1-cos-x-dx-




Question Number 174965 by cortano1 last updated on 15/Aug/22
      ∫ ((x+sin x)/(1+cos x)) dx =?
$$\:\:\:\:\:\:\int\:\frac{{x}+\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{cos}\:{x}}\:{dx}\:=? \\ $$
Answered by som(math1967) last updated on 15/Aug/22
∫(x/(2cos^2 (x/2)))dx+∫((2sin(x/2)cos(x/2))/(2cos^2 (x/2)))dx  (1/2)∫xsec^2 (x/2)dx +∫tan(x/2)dx  (x/2)∫sec^2 (x/2)dx−(1/2)∫{(dx/dx)∫sec^2 (x/2)dx}dx+∫tan(x/2)dx   xtan(x/2) +C
$$\int\frac{{x}}{\mathrm{2}{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{dx}+\int\frac{\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}}{\mathrm{2}{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int{xsec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}{dx}\:+\int{tan}\frac{{x}}{\mathrm{2}}{dx} \\ $$$$\frac{{x}}{\mathrm{2}}\int{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int\left\{\frac{{dx}}{{dx}}\int{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}{dx}\right\}{dx}+\int{tan}\frac{{x}}{\mathrm{2}}{dx}\: \\ $$$${xtan}\frac{{x}}{\mathrm{2}}\:+{C} \\ $$

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