Question Number 164290 by HongKing last updated on 15/Jan/22
Answered by Kamel last updated on 16/Jan/22
![Ω(n)=∫_0 ^π cot((x/2))Arctan(((nsin(x))/(1+ncos(x))))dx; u=((n−1)/(n+1)), n∈]−1;1]. =2∫_0 ^(+∞) (1/(t(1+t^2 )))Arctan((((1+u)t)/(1−ut^2 )))dt =2∫_0 ^(+∞) (1/(t(1+t^2 )))(Arctan(t)+Arctan(tu))dt =2∫_(−1) ^u ∫_0 ^(+∞) ((dtda)/((1+t^2 )(1+a^2 t^2 )))=2∫_(−1) ^u (1/(1−a^2 ))∫_0 ^(+∞) ((1/(1+t^2 ))−(a^2 /(1+a^2 t^2 )))dtda =2∫_(−1) ^u (1/(1−a^2 ))((π/2)+((aπ)/2))da=πLn((2/(1−u))) ∴ ∫_0 ^π cot((x/2))Arctan(((nsin(x))/(1+ncos(x))))dx=πLn(1+n)](https://www.tinkutara.com/question/Q164353.png)
$$ \\ $$$$\left.\Omega\left.\left({n}\right)=\int_{\mathrm{0}} ^{\pi} {cot}\left(\frac{{x}}{\mathrm{2}}\right){Arctan}\left(\frac{{nsin}\left({x}\right)}{\mathrm{1}+{ncos}\left({x}\right)}\right){dx};\:{u}=\frac{{n}−\mathrm{1}}{{n}+\mathrm{1}},\:{n}\in\right]−\mathrm{1};\mathrm{1}\right]. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{1}}{{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{Arctan}\left(\frac{\left(\mathrm{1}+{u}\right){t}}{\mathrm{1}−{ut}^{\mathrm{2}} }\right){dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{+\infty} \frac{\mathrm{1}}{{t}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\left({Arctan}\left({t}\right)+{Arctan}\left({tu}\right)\right){dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\int_{−\mathrm{1}} ^{{u}} \int_{\mathrm{0}} ^{+\infty} \frac{{dtda}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+{a}^{\mathrm{2}} {t}^{\mathrm{2}} \right)}=\mathrm{2}\int_{−\mathrm{1}} ^{{u}} \frac{\mathrm{1}}{\mathrm{1}−{a}^{\mathrm{2}} }\int_{\mathrm{0}} ^{+\infty} \left(\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }−\frac{{a}^{\mathrm{2}} }{\mathrm{1}+{a}^{\mathrm{2}} {t}^{\mathrm{2}} }\right){dtda} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\int_{−\mathrm{1}} ^{{u}} \frac{\mathrm{1}}{\mathrm{1}−{a}^{\mathrm{2}} }\left(\frac{\pi}{\mathrm{2}}+\frac{{a}\pi}{\mathrm{2}}\right){da}=\pi{Ln}\left(\frac{\mathrm{2}}{\mathrm{1}−{u}}\right) \\ $$$$\:\:\:\therefore\:\:\int_{\mathrm{0}} ^{\pi} {cot}\left(\frac{{x}}{\mathrm{2}}\right){Arctan}\left(\frac{{nsin}\left({x}\right)}{\mathrm{1}+{ncos}\left({x}\right)}\right){dx}=\pi{Ln}\left(\mathrm{1}+{n}\right) \\ $$
Answered by mnjuly1970 last updated on 16/Jan/22
$$\:\:{n}=\mathrm{1} \\ $$$$\:\:\:\:\int_{\mathrm{0}} ^{\:\pi} {cot}\left(\frac{{x}}{\mathrm{2}}\:\right){arctan}\left(\frac{{sin}\left({x}\right)}{\mathrm{1}+{cos}\left({x}\right)}\right){dx} \\ $$$$\:\:\:\:=\:\int_{\mathrm{0}} ^{\:\pi} \:{cot}\left(\frac{{x}}{\mathrm{2}}\right).{arctan}\left({tan}\left(\frac{{x}}{\mathrm{2}}\right)\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\:}\int_{\mathrm{0}} ^{\:\pi} {x}.{cot}\left(\frac{{x}}{\mathrm{2}}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {t}.{cot}\left({t}\right){dt} \\ $$$$\:\:=−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left(\:{sin}\left({t}\right)\right){dt}=\:\pi{ln}\left(\mathrm{2}\right) \\ $$