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Question Number 38210 by prof Abdo imad last updated on 22/Jun/18

let f(a)= ∫_0 ^π (dθ/(a +sin^2 θ)) (a from R) 1) find f(a) 2)calculate g(a)= ∫_0 ^π (dθ/((a+sin^2 θ)^2 )) 3)calculate ∫_0 ^π (dθ/(1+sin^2 θ)) and ∫_0 ^π (dθ/(2+sin^2 θ)) 4) calculate ∫_0 ^π (dθ/((3 +sin^2 θ)^2 )) .

$${let}\:{f}\left({a}\right)=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{d}\theta}{{a}\:+{sin}^{\mathrm{2}} \theta}\:\:\:\left({a}\:{from}\:{R}\right) \\ $$$$\left.\mathrm{1}\right)\:{find}\:{f}\left({a}\right) \\ $$$$\left.\mathrm{2}\right){calculate}\:{g}\left({a}\right)=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{d}\theta}{\left({a}+{sin}^{\mathrm{2}} \theta\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right){calculate}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{d}\theta}{\mathrm{1}+{sin}^{\mathrm{2}} \theta}\:{and}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{d}\theta}{\mathrm{2}+{sin}^{\mathrm{2}} \theta} \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{d}\theta}{\left(\mathrm{3}\:+{sin}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\:. \\ $$

Commented by math khazana by abdo last updated on 23/Jun/18

we have f(a) =∫_0 ^π (dθ/(a +((1−cos(2θ))/2))) = ∫_0 ^π ((2dθ)/(2a+1 −cos(2θ))) =_(2θ=t) ∫_0 ^(2π) (2/(2a+1−cost)) (dt/2) = ∫_0 ^(2π) (dt/(2a+1 −cost)) changement e^(it) =z give f(a) = ∫_(∣z∣=1) (1/(2a+1−((z+z^(−1) )/2))) (dz/(iz)) = ∫_(∣z∣=1) ((−2idz)/(z{ 4a+2−z−z^(−1) })) =∫_(∣z∣=1) ((−2idz)/((4a+2)z−z^2 −1)) =∫_(∣z∣=1) ((2idz)/(z^2 −(4a+2)z +1)) let ϕ(z)= ((2i)/(z^2 −(4a+2)z +1)) .poles of ϕ? Δ^′ =(2a+1)^2 −1=4a^2 +4a =4a(a+1) case 1 a(a+1)>0 ⇒z_1 = 2a+1 +2(√(a^2 +a)) z_2 =2a+1−2(√(a^2 +a)) ∣z_2 ∣ −1 =2a−2(√(a^2 +a))<0 and its clear that ∣z_1 ∣−1>0(to elominate from residus) ∫_(∣z∣=1) ϕ(z)dz =2iπ Res(ϕ,z_2 ) =((2i)/(z_2 −z_1 )) 2iπ =((−4π)/(−4(√(a^2 +a)))) = (π/(√(a^2 +a))) ⇒ f(a)=(π/(√(a^2 +a))) case 2 a(a+1)<0 ⇔−1<a<0 ⇒Δ^, =−4(−a(a+1)) =(2i(√(−a^2 −a)))^2 z_1 = 2a+1 +2i(√(−a^2 −a)) z_2 =2a+1−2i(√(−a^2 −a)) ∣z_1 ∣=(√( (2a+1)^2 +4(−a^2 −a))) =(√(4a^2 +4a +1−4a^2 −4a))=1 ∣z_2 ∣=1 ⇒ ∫_(∣z∣=1) ϕ(z)dz =2iπ {Res(ϕ,z_1 )+Res(ϕ,z_2 )} Res(ϕ,z_1 ) = ((2i)/(z_1 −z_2 )) Res(ϕ,z_2 ) = ((2i)/(z_2 −z_1 )) ⇒ ∫_(∣z∣=1) ϕ(z)dz=0 ⇒f(a)=0

$${we}\:{have}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{d}\theta}{{a}\:+\frac{\mathrm{1}−{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}} \\ $$$$=\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{\mathrm{2}{d}\theta}{\mathrm{2}{a}+\mathrm{1}\:−{cos}\left(\mathrm{2}\theta\right)}\:=_{\mathrm{2}\theta={t}} \:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{\mathrm{2}}{\mathrm{2}{a}+\mathrm{1}−{cost}}\:\frac{{dt}}{\mathrm{2}} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{{dt}}{\mathrm{2}{a}+\mathrm{1}\:−{cost}}\:\:{changement}\:{e}^{{it}} ={z}\:{give} \\ $$$${f}\left({a}\right)\:=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}{a}+\mathrm{1}−\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}}\:\frac{{dz}}{{iz}} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{−\mathrm{2}{idz}}{{z}\left\{\:\:\mathrm{4}{a}+\mathrm{2}−{z}−{z}^{−\mathrm{1}} \right\}} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{−\mathrm{2}{idz}}{\left(\mathrm{4}{a}+\mathrm{2}\right){z}−{z}^{\mathrm{2}} −\mathrm{1}}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{\mathrm{2}{idz}}{{z}^{\mathrm{2}} −\left(\mathrm{4}{a}+\mathrm{2}\right){z}\:+\mathrm{1}} \\ $$$${let}\:\varphi\left({z}\right)=\:\frac{\mathrm{2}{i}}{{z}^{\mathrm{2}} \:−\left(\mathrm{4}{a}+\mathrm{2}\right){z}\:+\mathrm{1}}\:.{poles}\:{of}\:\varphi? \\ $$$$\Delta^{'} \:=\left(\mathrm{2}{a}+\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{1}=\mathrm{4}{a}^{\mathrm{2}} \:+\mathrm{4}{a}\:=\mathrm{4}{a}\left({a}+\mathrm{1}\right) \\ $$$${case}\:\mathrm{1}\:\:{a}\left({a}+\mathrm{1}\right)>\mathrm{0}\:\Rightarrow{z}_{\mathrm{1}} =\:\mathrm{2}{a}+\mathrm{1}\:+\mathrm{2}\sqrt{{a}^{\mathrm{2}} \:+{a}} \\ $$$${z}_{\mathrm{2}} =\mathrm{2}{a}+\mathrm{1}−\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{a}} \\ $$$$\mid{z}_{\mathrm{2}} \mid\:−\mathrm{1}\:=\mathrm{2}{a}−\mathrm{2}\sqrt{{a}^{\mathrm{2}} \:+{a}}<\mathrm{0}\:{and}\:{its}\:{clear}\:{that} \\ $$$$\mid{z}_{\mathrm{1}} \mid−\mathrm{1}>\mathrm{0}\left({to}\:{elominate}\:{from}\:{residus}\right) \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{z}_{\mathrm{2}} \right)\:=\frac{\mathrm{2}{i}}{{z}_{\mathrm{2}} −{z}_{\mathrm{1}} }\:\mathrm{2}{i}\pi \\ $$$$=\frac{−\mathrm{4}\pi}{−\mathrm{4}\sqrt{{a}^{\mathrm{2}} +{a}}}\:=\:\frac{\pi}{\sqrt{{a}^{\mathrm{2}} +{a}}}\:\Rightarrow \\ $$$${f}\left({a}\right)=\frac{\pi}{\sqrt{{a}^{\mathrm{2}} \:+{a}}} \\ $$$${case}\:\mathrm{2}\:{a}\left({a}+\mathrm{1}\right)<\mathrm{0}\:\:\Leftrightarrow−\mathrm{1}<{a}<\mathrm{0}\:\Rightarrow\Delta^{,} =−\mathrm{4}\left(−{a}\left({a}+\mathrm{1}\right)\right) \\ $$$$=\left(\mathrm{2}{i}\sqrt{−{a}^{\mathrm{2}} −{a}}\right)^{\mathrm{2}} \\ $$$${z}_{\mathrm{1}} =\:\mathrm{2}{a}+\mathrm{1}\:+\mathrm{2}{i}\sqrt{−{a}^{\mathrm{2}} −{a}} \\ $$$${z}_{\mathrm{2}} =\mathrm{2}{a}+\mathrm{1}−\mathrm{2}{i}\sqrt{−{a}^{\mathrm{2}} −{a}} \\ $$$$\mid{z}_{\mathrm{1}} \mid=\sqrt{\:\left(\mathrm{2}{a}+\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{4}\left(−{a}^{\mathrm{2}} −{a}\right)} \\ $$$$=\sqrt{\mathrm{4}{a}^{\mathrm{2}} +\mathrm{4}{a}\:+\mathrm{1}−\mathrm{4}{a}^{\mathrm{2}} −\mathrm{4}{a}}=\mathrm{1} \\ $$$$\mid{z}_{\mathrm{2}} \mid=\mathrm{1}\:\Rightarrow \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{{Res}\left(\varphi,{z}_{\mathrm{1}} \right)+{Res}\left(\varphi,{z}_{\mathrm{2}} \right)\right\} \\ $$$${Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:=\:\frac{\mathrm{2}{i}}{{z}_{\mathrm{1}} −{z}_{\mathrm{2}} } \\ $$$${Res}\left(\varphi,{z}_{\mathrm{2}} \right)\:=\:\frac{\mathrm{2}{i}}{{z}_{\mathrm{2}} −{z}_{\mathrm{1}} }\:\Rightarrow\:\int_{\mid{z}\mid=\mathrm{1}} \:\varphi\left({z}\right){dz}=\mathrm{0}\:\Rightarrow{f}\left({a}\right)=\mathrm{0} \\ $$

Commented by math khazana by abdo last updated on 23/Jun/18

2) we have f^′ (a) = ∫_0 ^π (∂/∂a)( (1/(a+sin^2 θ)))dθ =−∫_0 ^π (dθ/((a+sin^2 θ)^2 )) =−g(a) ⇒ g(a) =−f^′ (a) if a(a+1)<0 ⇒g(a)=0 if a(a+1)>0 f(a)= (π/(√(a^2 +a))) ⇒ f^′ (a) = π{(a^2 +a)^(−(1/2)) }^′ =−(π/2)(2a+1)(a^2 +a)^(−(3/2)) = ((−π(2a+1))/(2(a^2 +a)(√(a^2 +a)))) ⇒ g(a) = (((2a+1)π)/(2(a^2 +a)(√(a^2 +a)))) .

$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}^{'} \left({a}\right)\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{\partial}{\partial{a}}\left(\:\frac{\mathrm{1}}{{a}+{sin}^{\mathrm{2}} \theta}\right){d}\theta \\ $$$$=−\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\:\frac{{d}\theta}{\left({a}+{sin}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\:=−{g}\left({a}\right)\:\Rightarrow \\ $$$${g}\left({a}\right)\:=−{f}^{'} \left({a}\right)\: \\ $$$${if}\:\:{a}\left({a}+\mathrm{1}\right)<\mathrm{0}\:\Rightarrow{g}\left({a}\right)=\mathrm{0} \\ $$$${if}\:{a}\left({a}+\mathrm{1}\right)>\mathrm{0}\:{f}\left({a}\right)=\:\frac{\pi}{\sqrt{{a}^{\mathrm{2}} \:+{a}}}\:\Rightarrow \\ $$$${f}^{'} \left({a}\right)\:=\:\pi\left\{\left({a}^{\mathrm{2}} +{a}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \right\}^{'} \:=−\frac{\pi}{\mathrm{2}}\left(\mathrm{2}{a}+\mathrm{1}\right)\left({a}^{\mathrm{2}} \:+{a}\right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$=\:\frac{−\pi\left(\mathrm{2}{a}+\mathrm{1}\right)}{\mathrm{2}\left({a}^{\mathrm{2}} +{a}\right)\sqrt{{a}^{\mathrm{2}} \:+{a}}}\:\Rightarrow \\ $$$${g}\left({a}\right)\:=\:\frac{\left(\mathrm{2}{a}+\mathrm{1}\right)\pi}{\mathrm{2}\left({a}^{\mathrm{2}} \:+{a}\right)\sqrt{{a}^{\mathrm{2}} \:+{a}}}\:. \\ $$$$ \\ $$

Commented by math khazana by abdo last updated on 23/Jun/18

3) we have proved that f(a)=(π/(√(a^2 +a))) if a(a+1)>0 so ∫_0 ^π (dθ/(1+sin^2 θ)) =f(1) = (π/(√2)) ∫_0 ^π (dθ/(2+sin^2 θ)) = f(2)= (π/(√6)) (/)

$$\left.\mathrm{3}\right)\:{we}\:{have}\:{proved}\:{that}\:\:{f}\left({a}\right)=\frac{\pi}{\sqrt{{a}^{\mathrm{2}} \:+{a}}}\:{if}\:{a}\left({a}+\mathrm{1}\right)>\mathrm{0} \\ $$$${so}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{d}\theta}{\mathrm{1}+{sin}^{\mathrm{2}} \theta}\:={f}\left(\mathrm{1}\right)\:=\:\frac{\pi}{\sqrt{\mathrm{2}}} \\ $$$$\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{d}\theta}{\mathrm{2}+{sin}^{\mathrm{2}} \theta}\:=\:{f}\left(\mathrm{2}\right)=\:\frac{\pi}{\sqrt{\mathrm{6}}} \\ $$$$ \\ $$$$\frac{}{} \\ $$

Commented by math khazana by abdo last updated on 23/Jun/18

4) ∫_0 ^π (dθ/((3+sin^2 θ)^2 )) =g(3)= ((7π)/(24(√(12)))) =((7π)/(48(√3))) .

$$\left.\mathrm{4}\right)\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\frac{{d}\theta}{\left(\mathrm{3}+{sin}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\:={g}\left(\mathrm{3}\right)=\:\frac{\mathrm{7}\pi}{\mathrm{24}\sqrt{\mathrm{12}}}\:=\frac{\mathrm{7}\pi}{\mathrm{48}\sqrt{\mathrm{3}}}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jun/18

∫_0 ^Π (dθ/(a+sin^2 θ)) ∫((sec^2 θ)/(asec^2 θ+tan^2 θ))dθ t=tanθ dt=sec^2 θdθ ∫(dt/(a+at^2 +t^2 )) ∫(dt/(a+t^2 (1+a))) (1/(1+a))∫(dt/((a/(1+a))+t^2 )) (1/(1+a))×(1/(((√a) )/(√(1+a))))tan^(−1) ((t/((√a)/(√(1+a))))) =2∣(1/((√a) ×(√(1+a)) ))×tan^(−1) (((tanθ)/(((√a) )/((√(1+a)) ))))∣_0 ^(Π/2) =2×(1/((√a) ×(√(1+a))))×(Π/2) =(Π/(√(a+a^2 ))) Ans ∫_0 ^Π f(θ)dθ=2∫_0 ^(Π/2) f(θ)dθ when f(Π−θ)=f(θ)

$$\int_{\mathrm{0}} ^{\Pi} \frac{{d}\theta}{{a}+{sin}^{\mathrm{2}} \theta} \\ $$$$\int\frac{{sec}^{\mathrm{2}} \theta}{{asec}^{\mathrm{2}} \theta+{tan}^{\mathrm{2}} \theta}{d}\theta \\ $$$${t}={tan}\theta\:\:{dt}={sec}^{\mathrm{2}} \theta{d}\theta \\ $$$$\int\frac{{dt}}{{a}+{at}^{\mathrm{2}} +{t}^{\mathrm{2}} } \\ $$$$\int\frac{{dt}}{{a}+{t}^{\mathrm{2}} \left(\mathrm{1}+{a}\right)} \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{a}}\int\frac{{dt}}{\frac{{a}}{\mathrm{1}+{a}}+{t}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{a}}×\frac{\mathrm{1}}{\frac{\sqrt{{a}}\:}{\sqrt{\mathrm{1}+{a}}}}{tan}^{−\mathrm{1}} \left(\frac{{t}}{\frac{\sqrt{{a}}}{\sqrt{\mathrm{1}+{a}}}}\right) \\ $$$$=\mathrm{2}\mid\frac{\mathrm{1}}{\sqrt{{a}}\:×\sqrt{\mathrm{1}+{a}}\:}×{tan}^{−\mathrm{1}} \left(\frac{{tan}\theta}{\frac{\sqrt{{a}}\:}{\sqrt{\mathrm{1}+{a}}\:}}\right)\mid_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} \: \\ $$$$=\mathrm{2}×\frac{\mathrm{1}}{\sqrt{{a}}\:×\sqrt{\mathrm{1}+{a}}}×\frac{\Pi}{\mathrm{2}} \\ $$$$=\frac{\Pi}{\sqrt{{a}+{a}^{\mathrm{2}} }}\:\:\mathscr{A}{ns} \\ $$$$ \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\Pi} {f}\left(\theta\right){d}\theta=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\Pi}{\mathrm{2}}} {f}\left(\theta\right){d}\theta\:\:{when}\:{f}\left(\Pi−\theta\right)={f}\left(\theta\right) \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Jun/18

g(a)=∫_0 ^Π (dθ/((a+sin^2 θ)^2 )) f(a)=∫_0 ^Π (dθ/(a+sin^2 θ)) ((df(a))/da)=∫_0 ^Π (∂/∂a)(1/((a+sin^2 θ)))dθ =∫_0 ^Π ((−1)/((a+sin^2 θ)^2 ))dθ so((df(a))/da)=−g(a) g(a)=−(d/da)((Π/((√(a+a^2 )) ))) g(a)=−Π×((−1)/(2(a+a^2 )^(3/2) ))×(1+2a) g(a)=((Π(1+2a))/(2(a+a^2 )^(3/2) )) ANS i have proved ∫_0 ^Π (dθ/(a+sin^2 θ))=(Π/(√(a+a^2 )))

$${g}\left({a}\right)=\int_{\mathrm{0}} ^{\Pi} \frac{{d}\theta}{\left({a}+{sin}^{\mathrm{2}} \theta\right)^{\mathrm{2}} } \\ $$$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\Pi} \frac{{d}\theta}{{a}+{sin}^{\mathrm{2}} \theta} \\ $$$$\frac{{df}\left({a}\right)}{{da}}=\int_{\mathrm{0}} ^{\Pi} \frac{\partial}{\partial{a}}\frac{\mathrm{1}}{\left({a}+{sin}^{\mathrm{2}} \theta\right)}{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\Pi} \frac{−\mathrm{1}}{\left({a}+{sin}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }{d}\theta \\ $$$${so}\frac{{df}\left({a}\right)}{{da}}=−{g}\left({a}\right) \\ $$$${g}\left({a}\right)=−\frac{{d}}{{da}}\left(\frac{\Pi}{\sqrt{{a}+{a}^{\mathrm{2}} }\:}\right) \\ $$$${g}\left({a}\right)=−\Pi×\frac{−\mathrm{1}}{\mathrm{2}\left({a}+{a}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }×\left(\mathrm{1}+\mathrm{2}{a}\right) \\ $$$${g}\left({a}\right)=\frac{\Pi\left(\mathrm{1}+\mathrm{2}{a}\right)}{\mathrm{2}\left({a}+{a}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\:\:{ANS} \\ $$$${i}\:{have}\:{proved}\:\int_{\mathrm{0}} ^{\Pi} \frac{{d}\theta}{{a}+{sin}^{\mathrm{2}} \theta}=\frac{\Pi}{\sqrt{{a}+{a}^{\mathrm{2}} }} \\ $$$$ \\ $$

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