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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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