Question Number 212239 by Spillover last updated on 07/Oct/24
Answered by Ghisom last updated on 08/Oct/24
![∫((e^(−x/2) (√(1−sin x)))/(1+cos x))dx= [t=tan (x/2)] =∫((e^(−arctan t) (t−1))/( (√(t^2 +1))))dt= =∫ ((e^(−arctan t) t)/( (√(t^2 +1))))dt−∫ (e^(−arctan t) /( (√(t^2 +1))))dt= [by parts] =e^(−arctan t) (√(t^2 +1))+∫ (e^(−arctan t) /( (√(t^2 +1))))dt−∫ (e^(−arctan t) /( (√(t^2 +1))))dt= =e^(−arctan t) (√(t^2 +1))= =(e^(−x/2) /(cos (x/2)))+C](https://www.tinkutara.com/question/Q212241.png)
$$\int\frac{\mathrm{e}^{−{x}/\mathrm{2}} \sqrt{\mathrm{1}−\mathrm{sin}\:{x}}}{\mathrm{1}+\mathrm{cos}\:{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right] \\ $$$$=\int\frac{\mathrm{e}^{−\mathrm{arctan}\:{t}} \left({t}−\mathrm{1}\right)}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}{dt}= \\ $$$$=\int\:\frac{\mathrm{e}^{−\mathrm{arctan}\:{t}} {t}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}{dt}−\int\:\frac{\mathrm{e}^{−\mathrm{arctan}\:{t}} }{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{by}\:\mathrm{parts}\right] \\ $$$$=\mathrm{e}^{−\mathrm{arctan}\:{t}} \sqrt{{t}^{\mathrm{2}} +\mathrm{1}}+\int\:\frac{\mathrm{e}^{−\mathrm{arctan}\:{t}} }{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}{dt}−\int\:\frac{\mathrm{e}^{−\mathrm{arctan}\:{t}} }{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}{dt}= \\ $$$$=\mathrm{e}^{−\mathrm{arctan}\:{t}} \sqrt{{t}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=\frac{\mathrm{e}^{−{x}/\mathrm{2}} }{\mathrm{cos}\:\frac{{x}}{\mathrm{2}}}+{C} \\ $$