Question Number 228034 by Zouhir last updated on 13/Mar/26
Answered by Ghisom_ last updated on 13/Mar/26
$$\frac{\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}−\mathrm{cos}\:{x}}{\mathrm{2sin}\:{x}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right) \\ $$$$\int\frac{\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}{dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\int{dx}−\int\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:{dx}\right)= \\ $$$$=\frac{{x}}{\mathrm{2}}+\mathrm{ln}\:\mid\mathrm{cos}\:\frac{{x}}{\mathrm{2}}\mid\:+{C} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\frac{\pi}{\mathrm{4}}−\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{2}} \\ $$
Commented by Zouhir last updated on 13/Mar/26
$$\:{Excellent} \\ $$
Answered by TonyCWX last updated on 14/Mar/26
![t = tan ((x/2)) ⇒ dx = ((2dt)/(t^2 +1)) ⇒ I = ∫_0 ^1 (((1−t^2 )/(1+t^2 ))/(1+((2t)/(1+t^2 ))+((1−t^2 )/(1+t^2 ))))((2/(t^2 +1)))dt ⇒ I = ∫_0 ^1 (((1−t^2 )/(1+t^2 ))/((2t+2)/(1+t^2 )))((2/(t^2 +1)))dt ⇒ I = ∫_0 ^1 ((1−t^2 )/((t+1)(t^2 +1)))dt ⇒ I = ∫_0 ^1 ((1−t)/(t^2 +1))dt = ∫_0 ^1 (1/(t^2 +1))dt − ∫_0 ^1 (t/(t^2 +1))dt ⇒ I = [tan^(−1) (t)]_0 ^1 − [(1/2)ln(t^2 +1)]_0 ^1 ⇒ I = (π/4) − (1/2)ln (2)](https://www.tinkutara.com/question/Q228045.png)
$${t}\:=\:\mathrm{tan}\:\left(\frac{{x}}{\mathrm{2}}\right)\:\Rightarrow\:{dx}\:=\:\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{1}+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\left(\frac{\mathrm{2}}{{t}^{\mathrm{2}} +\mathrm{1}}\right){dt} \\ $$$$\Rightarrow\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{\frac{\mathrm{2}{t}+\mathrm{2}}{\mathrm{1}+{t}^{\mathrm{2}} }}\left(\frac{\mathrm{2}}{{t}^{\mathrm{2}} +\mathrm{1}}\right){dt} \\ $$$$\Rightarrow\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{t}^{\mathrm{2}} }{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt} \\ $$$$\Rightarrow\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{t}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}\:−\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}}{{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$\Rightarrow\:{I}\:=\:\left[\mathrm{tan}^{−\mathrm{1}} \left({t}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\:\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left({t}^{\mathrm{2}} +\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$\Rightarrow\:{I}\:=\:\frac{\pi}{\mathrm{4}}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{2}\right) \\ $$