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Question Number 30178 by abdo imad last updated on 17/Feb/18
calculate ∫_0 ^(π/2) (dx/(1+cosx cosθ)) with −π<θ<π .
$${calculate}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{dx}}{\mathrm{1}+{cosx}\:{cos}\theta}\:\:{with}\:−\pi<\theta<\pi\:. \\ $$
Commented by prof Abdo imad last updated on 22/Feb/18
let put cosθ=t I= ∫_0 ^(π/2) (dx/(1+tcosx)) the ch.tan((x/2))=u give I= ∫_0 ^∞ (1/(1+t ((1−u^2 )/(1+u^2 )))) ((2du)/(1+u^2 )) I = ∫_0 ^∞ ((2du)/(1+u^2 +t(1−u^2 )))du =∫_0 ^∞ ((2du)/(1+t +(1−t)u^2 ))= (2/(1+t))∫_0 ^∞ (du/(1+((1−t)/(1+t)) u^2 )) let tben use the ch. (√((1−t)/(1+t))) u= α ⇒ I= (2/(1+t))∫_0 ^∞ (1/(1+α^2 )) (√((1+t)/(1−t))) dα = (2/( (√(1−t^2 )))) ∫_0 ^∞ (dα/(1+α^2 )) = (π/( (√(1−t^2 )))) = (π/( (√(1−cos^2 θ))))⇒ I = (π/(∣sinθ∣)) if θ≠0 also we must study the case θ=0 ....
$${let}\:{put}\:{cos}\theta={t}\:{I}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{dx}}{\mathrm{1}+{tcosx}}\:{the}\:{ch}.{tan}\left(\frac{{x}}{\mathrm{2}}\right)={u} \\ $$$${give}\:\:{I}=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\mathrm{1}+{t}\:\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} \:+{t}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)}{du} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{2}{du}}{\mathrm{1}+{t}\:+\left(\mathrm{1}−{t}\right){u}^{\mathrm{2}} }=\:\frac{\mathrm{2}}{\mathrm{1}+{t}}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{du}}{\mathrm{1}+\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}\:{u}^{\mathrm{2}} } \\ $$$${let}\:{tben}\:{use}\:{the}\:{ch}.\:\sqrt{\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}}\:{u}=\:\alpha\:\Rightarrow \\ $$$${I}=\:\:\frac{\mathrm{2}}{\mathrm{1}+{t}}\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+\alpha^{\mathrm{2}} }\:\sqrt{\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}}\:\:{d}\alpha \\ $$$$=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{d}\alpha}{\mathrm{1}+\alpha^{\mathrm{2}} }\:=\:\frac{\pi}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:=\:\frac{\pi}{\:\sqrt{\mathrm{1}−{cos}^{\mathrm{2}} \theta}}\Rightarrow \\ $$$${I}\:=\:\frac{\pi}{\mid{sin}\theta\mid}\:\:{if}\:\theta\neq\mathrm{0}\:\:{also}\:{we}\:{must}\:{study}\:{the}\:{case}\:\theta=\mathrm{0} \\ $$$$…. \\ $$

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