Question Number 32712 by caravan msup abdo. last updated on 31/Mar/18
$${calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{dt}}{\mathrm{1}+{a}\:{cos}^{\mathrm{2}} {t}}\:. \\ $$
Commented by abdo imad last updated on 03/Apr/18
![let put F(a) = ∫_0 ^(π/2) (dt/(1+a cos^2 t)) F(a) = ∫_0 ^(π/2) (dt/(1+a ((1+cos(2t))/2))) =∫_0 ^(π/2) ((2dt)/(2+a +acos(2t))) =_(2t =u) ∫_0 ^π (2/(2+a +acos(u))) (du/2) = ∫_0 ^π (du/(2+a +acosu)) but the ch. tan((u/2))=x give F(a) = ∫_0 ^(+∞) (1/(2+a +a ((1−x^2 )/(1+x^2 )))) ((2dx)/(1+x^2 )) = ∫_0 ^∞ ((2dx)/((2+a)(1+x^2 ) +a(1−x^2 ))) = ∫_0 ^∞ ((2dx)/(2+a +(2+a)x^2 +a −ax^2 )) = ∫_0 ^∞ ((2dx)/( 2+2a +2x^2 )) = ∫_0 ^∞ (dx/(1+a +x^2 )) case 1 1+a >0 ⇒ F(a) =_(x=(√(1+a))u) ∫_0 ^∞ (((√(1+a))du)/((1+a)(1+u^2 ))) = (π/(2(√(1+a)))) case2 1+a<0 ⇒ F(a) = ∫_0 ^∞ (dx/(x^2 −((√(−(1+a))))^2 )) = ∫_0 ^∞ (dx/((x −α)(x+α))) (α=(√(−1−a))) = (1/(2α))∫_0 ^∞ ( (1/(x−α)) −(1/(x+α)))dx =(1/(2α)) [ln∣((x−α)/(x+α))∣]_0 ^(+∞) =0](Q32851.png)
$${let}\:{put}\:{F}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dt}}{\mathrm{1}+{a}\:{cos}^{\mathrm{2}} {t}} \\ $$$${F}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{dt}}{\mathrm{1}+{a}\:\frac{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{2}+{a}\:+{acos}\left(\mathrm{2}{t}\right)} \\ $$$$=_{\mathrm{2}{t}\:={u}} \:\:\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{\mathrm{2}}{\mathrm{2}+{a}\:+{acos}\left({u}\right)}\:\frac{{du}}{\mathrm{2}} \\ $$$$=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{du}}{\mathrm{2}+{a}\:+{acosu}}\:\:{but}\:{the}\:{ch}.\:{tan}\left(\frac{{u}}{\mathrm{2}}\right)={x}\:{give} \\ $$$${F}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\:\frac{\mathrm{1}}{\mathrm{2}+{a}\:+{a}\:\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }}\:\:\frac{\mathrm{2}{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{\mathrm{2}{dx}}{\left(\mathrm{2}+{a}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\:+{a}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{2}{dx}}{\mathrm{2}+{a}\:\:+\left(\mathrm{2}+{a}\right){x}^{\mathrm{2}} \:+{a}\:−{ax}^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{2}{dx}}{\:\mathrm{2}+\mathrm{2}{a}\:+\mathrm{2}{x}^{\mathrm{2}} }\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\mathrm{1}+{a}\:+{x}^{\mathrm{2}} } \\ $$$${case}\:\mathrm{1}\:\:\mathrm{1}+{a}\:>\mathrm{0}\:\:\Rightarrow\:{F}\left({a}\right)\:=_{{x}=\sqrt{\mathrm{1}+{a}}{u}} \:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\sqrt{\mathrm{1}+{a}}{du}}{\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{u}^{\mathrm{2}} \right)} \\ $$$$=\:\frac{\pi}{\mathrm{2}\sqrt{\mathrm{1}+{a}}} \\ $$$${case}\mathrm{2}\:\:\mathrm{1}+{a}<\mathrm{0}\:\Rightarrow\:{F}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:−\left(\sqrt{−\left(\mathrm{1}+{a}\right)}\right)^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left({x}\:−\alpha\right)\left({x}+\alpha\right)}\:\:\:\:\:\:\:\left(\alpha=\sqrt{−\mathrm{1}−{a}}\right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}\alpha}\int_{\mathrm{0}} ^{\infty} \:\left(\:\frac{\mathrm{1}}{{x}−\alpha}\:−\frac{\mathrm{1}}{{x}+\alpha}\right){dx}\:=\frac{\mathrm{1}}{\mathrm{2}\alpha}\:\left[{ln}\mid\frac{{x}−\alpha}{{x}+\alpha}\mid\right]_{\mathrm{0}} ^{+\infty} \:=\mathrm{0} \\ $$