Question Number 78271 by msup trace by abdo last updated on 15/Jan/20
$${calculate}\:{A}_{\theta} \:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dx}}{\mathrm{2}+{cos}\theta\:{sinx}} \\ $$ $$−\pi<\theta<\pi \\ $$
Commented bymathmax by abdo last updated on 19/Jan/20
![A(θ)=∫_0 ^(π/2) (dx/(2 +cosθ sinx)) ⇒A(θ)=_(tan((x/2))=t) ∫_0 ^(1 ) ((2dt)/((1+t^2 )(2+cosθ((2t)/(1+t^2 ))))) =∫_0 ^1 ((2dt)/(2+2t^2 +2t cosθ)) =∫_0 ^1 (dt/(t^2 +t cosθ +1)) =∫_0 ^1 (dt/(t^2 +2t((cosθ)/2) +((cos^2 θ)/4) +1−((cos^2 θ)/4))) =∫_0 ^1 (dt/((t+((cosθ)/2))^2 +((4−cos^2 θ)/4))) =_(t+((cosθ)/2)=((√(4−cos^2 θ))/2)u ⇒u=((2t+cosθ)/(√(4−cos^2 θ)))) =∫_((cosθ)/(√(4−cos^2 θ))) ^((2+cosθ)/(√(4−cos^2 θ))) (1/(((4−cos^2 θ)/4)(1+u^2 )))×((√(4−cos^2 θ))/2)du =2 ∫_((cosθ)/(√(4−cos^2 θ))) ^((2+cosθ)/(√(4−cos^2 θ))) (du/(1+u^2 )) =2 [arctan(u)]_((cosθ)/(√(4−cos^2 θ))) ^((2+cosθ)/(√(4−cos^2 θ))) =2{ arctan(((2+cosθ)/(√(4−cos^2 θ))))−arctan(((cosθ)/(√(4−cos^2 θ))))}.](Q78592.png)
$$\:{A}\left(\theta\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{dx}}{\mathrm{2}\:+{cos}\theta\:{sinx}}\:\Rightarrow{A}\left(\theta\right)=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\:\int_{\mathrm{0}} ^{\mathrm{1}\:} \:\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{2}+{cos}\theta\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)} \\ $$ $$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2}{dt}}{\mathrm{2}+\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}{t}\:{cos}\theta}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{{t}^{\mathrm{2}} \:+{t}\:{cos}\theta\:+\mathrm{1}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{2}{t}\frac{{cos}\theta}{\mathrm{2}}\:+\frac{{cos}^{\mathrm{2}} \theta}{\mathrm{4}}\:+\mathrm{1}−\frac{{cos}^{\mathrm{2}} \theta}{\mathrm{4}}} \\ $$ $$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{\left({t}+\frac{{cos}\theta}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{4}−{cos}^{\mathrm{2}} \theta}{\mathrm{4}}}\:=_{{t}+\frac{{cos}\theta}{\mathrm{2}}=\frac{\sqrt{\mathrm{4}−{cos}^{\mathrm{2}} \theta}}{\mathrm{2}}{u}\:\Rightarrow{u}=\frac{\mathrm{2}{t}+{cos}\theta}{\sqrt{\mathrm{4}−{cos}^{\mathrm{2}} \theta}}} \\ $$ $$=\int_{\frac{{cos}\theta}{\sqrt{\mathrm{4}−{cos}^{\mathrm{2}} \theta}}} ^{\frac{\mathrm{2}+{cos}\theta}{\sqrt{\mathrm{4}−{cos}^{\mathrm{2}} \theta}}} \:\:\:\:\:\frac{\mathrm{1}}{\frac{\mathrm{4}−{cos}^{\mathrm{2}} \theta}{\mathrm{4}}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}×\frac{\sqrt{\mathrm{4}−{cos}^{\mathrm{2}} \theta}}{\mathrm{2}}{du} \\ $$ $$=\mathrm{2}\:\int_{\frac{{cos}\theta}{\sqrt{\mathrm{4}−{cos}^{\mathrm{2}} \theta}}} ^{\frac{\mathrm{2}+{cos}\theta}{\sqrt{\mathrm{4}−{cos}^{\mathrm{2}} \theta}}} \:\:\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\mathrm{2}\:\left[{arctan}\left({u}\right)\right]_{\frac{{cos}\theta}{\sqrt{\mathrm{4}−{cos}^{\mathrm{2}} \theta}}} ^{\frac{\mathrm{2}+{cos}\theta}{\sqrt{\mathrm{4}−{cos}^{\mathrm{2}} \theta}}} \\ $$ $$=\mathrm{2}\left\{\:{arctan}\left(\frac{\mathrm{2}+{cos}\theta}{\sqrt{\mathrm{4}−{cos}^{\mathrm{2}} \theta}}\right)−{arctan}\left(\frac{{cos}\theta}{\sqrt{\mathrm{4}−{cos}^{\mathrm{2}} \theta}}\right)\right\}. \\ $$