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Question Number 37350 by math khazana by abdo last updated on 12/Jun/18

fond the value of ∫_0 ^(2π) (dt/((a +cost)^2 )) with a>1.

$${fond}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{{dt}}{\left({a}\:+{cost}\right)^{\mathrm{2}} }\:\:{with}\:{a}>\mathrm{1}. \\ $$ $$ \\ $$

Commented bymath khazana by abdo last updated on 12/Jun/18

I = ∫_0 ^π (dt/((a +cost)^2 )) + ∫_π ^(2π) (dt/((a +cost)^2 )) =K+H changement tan((t/2))=x give k = ∫_0 ^∞ (1/({a+((1−x^2 )/(1+x^2 ))}^2 )) ((2dx)/(1+x^2 )) =2 ∫_0 ^∞ ((1+x^2 )/({a +ax^(2 ) +1−x^2 }))dx =2∫_0 ^∞ ((1+x^2 )/({a+1 +(a−1)x^2 }^2 ))dx = (2/((a+1)^2 )) ∫_0 ^∞ ((1+x^2 )/({ 1+((a−1)/(a+1))x^2 }^2 ))dx =_((√((a−1)/(a+1)))x=u) (2/((a+1)^2 )) ∫_0 ^∞ ((1+((a+1)/(a−1))u^2 )/({ 1+u^2 }^2 )) (√((a+1)/(a−1)))du = (1/((a−1)(√((a+1)/(a−1))))) (2/((a+1)^2 )) ∫_0 ^∞ ((a−1 +(a+1)u^2 )/({1+u^2 }^2 )) du = (2/((a+1)^2 (√(a^2 −1)))) ∫_0 ^∞ ((a−1 +(a+1)u^2 )/({1+u^2 }^2 )) du but ∫_0 ^∞ ((a−1 +(a+1)u^2 )/((1+u^2 )^2 ))du = (a−1)∫_0 ^∞ (du/((1+u^2 )^2 )) +(a+1)∫_0 ^∞ ((1+u^2 −1)/((1+u^2 )^2 ))du =(a−1)∫_0 ^∞ (du/((1+u^2 )^2 )) +(a+1)(π/2) −(a+1)∫_0 ^∞ (du/((1+u^2 )^2 )) =−2 ∫_0 ^∞ (du/((1+u^2 )^2 )) +(a+1)(π/2) chang.u=tanθ −(π/2) +((aπ)/2) +(π/2) =((πa)/2) because ∫_0 ^(+∞) (du/((1+u^2 )^2 )) = ∫_0 ^(π/2) (((1+tan^2 θ)dθ)/((1+tan^2 θ)^2 )) =∫_0 ^(π/2) (dθ/(1+tan^2 θ)) =∫_0 ^(π/2) cos^2 θ dθ =(1/2) ∫_0 ^(π/2) (1+cos(2θ))dθ = (π/4) ⇒ K = (2/((a+1)^2 (√(a^2 −1)))) ((πa)/2) =((πa)/((a+1)^2 (√(a^2 −1)))) .

$${I}\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dt}}{\left({a}\:+{cost}\right)^{\mathrm{2}} }\:\:+\:\int_{\pi} ^{\mathrm{2}\pi} \:\:\:\frac{{dt}}{\left({a}\:+{cost}\right)^{\mathrm{2}} }\:={K}+{H} \\ $$ $${changement}\:\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)={x}\:{give}\: \\ $$ $${k}\:=\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left\{{a}+\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }\right\}^{\mathrm{2}} }\:\frac{\mathrm{2}{dx}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\left\{{a}\:+{ax}^{\mathrm{2}\:} +\mathrm{1}−{x}^{\mathrm{2}} \right\}}{dx} \\ $$ $$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\left\{{a}+\mathrm{1}\:+\left({a}−\mathrm{1}\right){x}^{\mathrm{2}} \right\}^{\mathrm{2}} }{dx} \\ $$ $$=\:\frac{\mathrm{2}}{\left({a}+\mathrm{1}\right)^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\left\{\:\mathrm{1}+\frac{{a}−\mathrm{1}}{{a}+\mathrm{1}}{x}^{\mathrm{2}} \right\}^{\mathrm{2}} }{dx} \\ $$ $$=_{\sqrt{\frac{{a}−\mathrm{1}}{{a}+\mathrm{1}}}{x}={u}} \:\:\:\frac{\mathrm{2}}{\left({a}+\mathrm{1}\right)^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}+\frac{{a}+\mathrm{1}}{{a}−\mathrm{1}}{u}^{\mathrm{2}} }{\left\{\:\mathrm{1}+{u}^{\mathrm{2}} \right\}^{\mathrm{2}} }\:\sqrt{\frac{{a}+\mathrm{1}}{{a}−\mathrm{1}}}{du} \\ $$ $$=\:\:\frac{\mathrm{1}}{\left({a}−\mathrm{1}\right)\sqrt{\frac{{a}+\mathrm{1}}{{a}−\mathrm{1}}}}\:\frac{\mathrm{2}}{\left({a}+\mathrm{1}\right)^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{a}−\mathrm{1}\:+\left({a}+\mathrm{1}\right){u}^{\mathrm{2}} }{\left\{\mathrm{1}+{u}^{\mathrm{2}} \right\}^{\mathrm{2}} }\:{du} \\ $$ $$=\:\frac{\mathrm{2}}{\left({a}+\mathrm{1}\right)^{\mathrm{2}} \sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{a}−\mathrm{1}\:+\left({a}+\mathrm{1}\right){u}^{\mathrm{2}} }{\left\{\mathrm{1}+{u}^{\mathrm{2}} \right\}^{\mathrm{2}} }\:{du}\:\:{but} \\ $$ $$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{a}−\mathrm{1}\:+\left({a}+\mathrm{1}\right){u}^{\mathrm{2}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }{du} \\ $$ $$=\:\left({a}−\mathrm{1}\right)\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\:+\left({a}+\mathrm{1}\right)\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}+{u}^{\mathrm{2}} −\mathrm{1}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }{du} \\ $$ $$=\left({a}−\mathrm{1}\right)\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\:+\left({a}+\mathrm{1}\right)\frac{\pi}{\mathrm{2}}\:−\left({a}+\mathrm{1}\right)\int_{\mathrm{0}} ^{\infty} \:\:\frac{{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$ $$=−\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\:+\left({a}+\mathrm{1}\right)\frac{\pi}{\mathrm{2}}\:\:{chang}.{u}={tan}\theta \\ $$ $$−\frac{\pi}{\mathrm{2}}\:+\frac{{a}\pi}{\mathrm{2}}\:+\frac{\pi}{\mathrm{2}}\:=\frac{\pi{a}}{\mathrm{2}}\:\:{because} \\ $$ $$\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{\mathrm{2}} } \\ $$ $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{d}\theta}{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cos}^{\mathrm{2}} \theta\:{d}\theta \\ $$ $$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)\right){d}\theta\:=\:\frac{\pi}{\mathrm{4}}\:\Rightarrow \\ $$ $${K}\:=\:\frac{\mathrm{2}}{\left({a}+\mathrm{1}\right)^{\mathrm{2}} \sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}\:\frac{\pi{a}}{\mathrm{2}}\:\:=\frac{\pi{a}}{\left({a}+\mathrm{1}\right)^{\mathrm{2}} \sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}\:. \\ $$ $$ \\ $$ $$ \\ $$

Commented bymath khazana by abdo last updated on 12/Jun/18

H =∫_π ^(2π) (dt/((a +cost)^2 )) chang. t =π +x give H = ∫_0 ^π (dx/((a−cosx)^2 )) =_(tan((x/2))=u) ∫_0 ^(+∞) (1/((a−((1−u^2 )/(1+u^2 )))^2 )) ((2du)/(1+u^2 )) =2∫_0 ^(+∞) ((1+u^2 )/({a +au^2 −1+u^2 }^2 ))du ...we follow the same method to find H....be continued..

$${H}\:=\int_{\pi} ^{\mathrm{2}\pi} \:\:\:\:\frac{{dt}}{\left({a}\:+{cost}\right)^{\mathrm{2}} }\:\:{chang}.\:{t}\:=\pi\:+{x}\:{give} \\ $$ $${H}\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\frac{{dx}}{\left({a}−{cosx}\right)^{\mathrm{2}} }\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={u}} \:\int_{\mathrm{0}} ^{+\infty} \:\:\:\:\frac{\mathrm{1}}{\left({a}−\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }\right)^{\mathrm{2}} }\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$ $$=\mathrm{2}\int_{\mathrm{0}} ^{+\infty} \:\:\:\:\:\frac{\mathrm{1}+{u}^{\mathrm{2}} }{\left\{{a}\:+{au}^{\mathrm{2}} −\mathrm{1}+{u}^{\mathrm{2}} \right\}^{\mathrm{2}} }{du}\:...{we}\:{follow}\:{the} \\ $$ $${same}\:{method}\:{to}\:{find}\:{H}....{be}\:{continued}.. \\ $$

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