Question Number 67526 by mathmax by abdo last updated on 28/Aug/19
$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{dx}}{\mathrm{3}+\mathrm{2}{sinx}\:+{cosx}} \\ $$
Commented by mathmax by abdo last updated on 31/Aug/19
![let I =∫_0 ^(2π) (dx/(3+2sinx +cosx)) ⇒I =∫_0 ^π (dx/(3+2sinx +cosx)) +∫_π ^(2π) (dx/(3+2sinx +cosx)) =H +K H =_(tan((x/2))=t) ∫_0 ^∞ (1/(3+2((2t)/(1+t^2 )) +((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) =∫_0 ^∞ ((2dt)/(3+3t^2 +4t +1−t^2 )) =∫_0 ^∞ ((2dt)/(2t^2 +4t+4)) =∫_0 ^∞ (dt/(t^2 +2t +2)) Δ^′ =1−2 =−1<0 ⇒ H =∫_0 ^∞ (dt/(t^2 +2t+1 +1)) =∫_0 ^∞ (dt/((t+1)^2 +1)) =_(t+1 =α) ∫_1 ^(+∞) (dα/(1+α^2 )) =[arctan(α)]_1 ^(+∞) =(π/2)−(π/4) =(π/4) K =_(x =π +t) ∫_0 ^π (dt/(3−2sint−cost)) =_(tan((t/2))=u) ∫_0 ^∞ ((2du)/((1+u^2 )(3−2((2u)/(1+u^2 ))−((1−u^2 )/(1+u^2 ))))) =∫_0 ^∞ ((2du)/(3+3u^2 −4u−1+u^2 )) =∫_0 ^∞ ((2du)/(4u^2 −4u+2)) =∫_0 ^∞ (du/(2u^2 −2u +1)) =(1/2)∫_0 ^∞ (du/(u^2 −u +(1/2))) =(1/2)∫_0 ^∞ (du/((u−(1/2))^2 +(3/4)+(1/2))) =(1/2)∫_0 ^∞ (du/((u−(1/2))^2 +(5/4))) =_(u−(1/2)=((√5)/2)α) (1/2)×(4/5)∫_(−(1/( (√5)))) ^(+∞) (1/(1+u^2 ))((√5)/2)dα =(1/( (√5)))[arctanα]_(−(1/( (√5)))) ^(+∞) =(1/( (√5))){(π/2) +arctan((1/( (√5))))} ⇒ I =(π/4) +(π/(2(√5))) +(1/( (√5))) arctan((1/( (√5)))).](https://www.tinkutara.com/question/Q67802.png)
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dx}}{\mathrm{3}+\mathrm{2}{sinx}\:+{cosx}}\:\Rightarrow{I}\:=\int_{\mathrm{0}} ^{\pi} \:\frac{{dx}}{\mathrm{3}+\mathrm{2}{sinx}\:+{cosx}}\:+\int_{\pi} ^{\mathrm{2}\pi} \:\frac{{dx}}{\mathrm{3}+\mathrm{2}{sinx}\:+{cosx}} \\ $$$$={H}\:+{K} \\ $$$${H}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{1}}{\mathrm{3}+\mathrm{2}\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\:+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{dt}}{\mathrm{3}+\mathrm{3}{t}^{\mathrm{2}} +\mathrm{4}{t}\:+\mathrm{1}−{t}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{dt}}{\mathrm{2}{t}^{\mathrm{2}} +\mathrm{4}{t}+\mathrm{4}}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{2}{t}\:+\mathrm{2}} \\ $$$$\Delta^{'} =\mathrm{1}−\mathrm{2}\:=−\mathrm{1}<\mathrm{0}\:\Rightarrow \\ $$$${H}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{2}{t}+\mathrm{1}\:+\mathrm{1}}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{\left({t}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}\:=_{{t}+\mathrm{1}\:=\alpha} \:\:\:\int_{\mathrm{1}} ^{+\infty} \:\frac{{d}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} } \\ $$$$=\left[{arctan}\left(\alpha\right)\right]_{\mathrm{1}} ^{+\infty} \:=\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\:=\frac{\pi}{\mathrm{4}} \\ $$$${K}\:=_{{x}\:=\pi\:+{t}} \:\:\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\frac{{dt}}{\mathrm{3}−\mathrm{2}{sint}−{cost}}\:=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{2}{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{3}−\mathrm{2}\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }−\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }\right)} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{2}{du}}{\mathrm{3}+\mathrm{3}{u}^{\mathrm{2}} −\mathrm{4}{u}−\mathrm{1}+{u}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{du}}{\mathrm{4}{u}^{\mathrm{2}} −\mathrm{4}{u}+\mathrm{2}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{du}}{\mathrm{2}{u}^{\mathrm{2}} −\mathrm{2}{u}\:+\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{du}}{{u}^{\mathrm{2}} −{u}\:+\frac{\mathrm{1}}{\mathrm{2}}}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{du}}{\left({u}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{du}}{\left({u}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{4}}}\:=_{{u}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\alpha} \:\:\:\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{4}}{\mathrm{5}}\int_{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}} ^{+\infty} \:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}{d}\alpha \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left[{arctan}\alpha\right]_{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}} ^{+\infty} \:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\left\{\frac{\pi}{\mathrm{2}}\:+{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\right)\right\}\:\Rightarrow \\ $$$${I}\:=\frac{\pi}{\mathrm{4}}\:+\frac{\pi}{\mathrm{2}\sqrt{\mathrm{5}}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\:{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\right). \\ $$