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Question Number 187898 by cortano12 last updated on 23/Feb/23
∫ ((1−cos x)/(cos x+sin x−1)) dx=?
$$\:\:\int\:\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}−\mathrm{1}}\:\mathrm{dx}=? \\ $$
Answered by horsebrand11 last updated on 24/Feb/23
=(1/2)∫ ((2−2cos x)/(cos x+sin −1)) dx =(1/2)∫ (((1−cos x+sin x)−(cos x+sin x−1))/(cos x+sin x−1)) dx =−(1/2)x−(1/2)∫((sin (1/2)x+cos (1/2)x)/(sin (1/2)x−cos (1/2)x)) dx =−(1/2)x−(1/2)∫ ((d(sin (1/2)x−cos (1/2)x))/(sin (1/2)x−cos (1/2)x)) dx =−(1/2)x−(1/2)ln ∣sin (1/2)x−cos (1/2)x∣+c
$$\:=\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{2}−\mathrm{2cos}\:{x}}{\mathrm{cos}\:{x}+\mathrm{sin}\:−\mathrm{1}}\:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\left(\mathrm{1}−\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right)−\left(\mathrm{cos}\:{x}+\mathrm{sin}\:{x}−\mathrm{1}\right)}{\mathrm{cos}\:{x}+\mathrm{sin}\:{x}−\mathrm{1}}\:{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{x}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}+\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{x}}{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}−\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{x}}\:{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{x}−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{d}\left(\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}−\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{x}\right)}{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}−\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{x}}\:{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{x}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}−\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{x}\mid+{c}\: \\ $$
Commented by Ar Brandon last updated on 24/Feb/23
=−(1/2)x−∫((d(sin (1/2)x−cos (1/2)x))/(sin (1/2)x−cos (1/2)x))dx
$$=−\frac{\mathrm{1}}{\mathrm{2}}{x}−\int\frac{{d}\left(\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}−\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{x}\right)}{\mathrm{sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}−\mathrm{cos}\:\frac{\mathrm{1}}{\mathrm{2}}{x}}{dx} \\ $$
Answered by Ar Brandon last updated on 23/Feb/23
I=∫((1−cosx)/(cosx+sinx−1))dx , t=tan(x/2) =∫((1−((1−t^2 )/(1+t^2 )))/(((1−t^2 )/(1+t^2 ))+((2t)/(1+t^2 ))−1))∙(2/(1+t^2 ))dt=4∫(t^2 /((2t−2t^2 )(1+t^2 )))dt =−2∫(t/((t−1)(t^2 +1)))=2∫(((t−1)/(2(t^2 +1)))−(1/(2(t−1))))dt =((ln(t^2 +1))/2)−arctan(t)−ln∣t−1∣+C =(1/2)ln(tan^2 ((x/2))+1)−(x/2)−ln∣tan((x/2))−1∣+C
$${I}=\int\frac{\mathrm{1}−\mathrm{cos}{x}}{\mathrm{cos}{x}+\mathrm{sin}{x}−\mathrm{1}}{dx}\:,\:{t}=\mathrm{tan}\frac{{x}}{\mathrm{2}} \\ $$$$\:\:=\int\frac{\mathrm{1}−\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }−\mathrm{1}}\centerdot\frac{\mathrm{2}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\mathrm{4}\int\frac{{t}^{\mathrm{2}} }{\left(\mathrm{2}{t}−\mathrm{2}{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt} \\ $$$$\:\:=−\mathrm{2}\int\frac{{t}}{\left({t}−\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}=\mathrm{2}\int\left(\frac{{t}−\mathrm{1}}{\mathrm{2}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{2}\left({t}−\mathrm{1}\right)}\right){dt} \\ $$$$\:\:=\frac{\mathrm{ln}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{2}}−\mathrm{arctan}\left({t}\right)−\mathrm{ln}\mid{t}−\mathrm{1}\mid+{C} \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{1}\right)−\frac{{x}}{\mathrm{2}}−\mathrm{ln}\mid\mathrm{tan}\left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{1}\mid+{C} \\ $$

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