∫_0 ^(2π) (dx/((1−ksinx)^2 ))


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Question Number 143399 by Ar Brandon last updated on 13/Jun/21

$\int_{0}^{2 π} \frac{dx}{{(1 - ksinx)}^{2}}$
	Answered by Ar Brandon last updated on 13/Jun/21

	$I = \int_{0}^{2 π} \frac{dx}{{(1 - ksinx)}^{2}} = 2 \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{dx}{{(1 - ksinx)}^{2}} t = \tan \frac{x}{2}$ $= 2 \int_{- 1}^{1} \frac{1}{{(1 - k \frac{2 t}{1 + t^{2}})}^{2}} \cdot \frac{2 dt}{1 + t^{2}} = 4 \int_{- 1}^{1} \frac{t^{2} + 1}{{(t^{2} - 2 kt + 1)}^{2}} dt$ $= 4 {[\frac{at + b}{t^{2} - 2 kt + 1} + \int \frac{ct + d}{t^{2} - 2 kt + 1} dt]}_{- 1}^{1}$ $= 4 \int [\frac{a (t^{2} - 2 kt + 1) - (at + b) (2 t - 2 k)}{{(t^{2} + 2 kt + 1)}^{2}} + \frac{ct + d}{t^{2} - 2 kt + 1}] dt$ $c = 0, a - 2 a + d = 1 \Rightarrow d = a + 1, - 2 ka + 2 ka - 2 b - 2 kd = 0 \Rightarrow b = - kd$ $a + 2 bk + d = 1 \Rightarrow d + bk = 1 \Rightarrow d (1 - k^{2}) = 1 \Rightarrow d = \frac{1}{1 - k^{2}}, b = \frac{k}{k^{2} - 1}$ $\Rightarrow a = \frac{1}{1 - k^{2}} - 1 = \frac{1 + k^{2} - 1}{1 - k^{2}} = \frac{k^{2}}{1 - k^{2}}$ $I = 4 {[\frac{\frac{k^{2}}{1 - k^{2}} t + \frac{k}{k^{2} - 1}}{t^{2} - 2 kt + 1} + \frac{1}{1 - k^{2}} \int \frac{dt}{t^{2} - 2 kt + 1}]}_{- 1}^{1}$ $= \frac{4}{2 - 2 k} (\frac{k^{2}}{1 - k^{2}} + \frac{1}{k^{2} - 1}) - \frac{4}{2 + 2 k} (\frac{k}{k^{2} - 1} - \frac{k^{2}}{1 - k^{2}}) + \frac{4}{1 - k^{2}} \int \frac{dt}{t^{2} - 2 kt + 1}$ $\int \frac{dt}{t^{2} - 2 kt + 1} = \int \frac{dt}{{(t - k)}^{2} + 1 - k^{2}} = \frac{1}{\sqrt{1 - k^{2}}} \arctan (\frac{t - k}{\sqrt{1 - k^{2}}}), ∣ k ∣< 1$ $\int \frac{dt}{t^{2} - 2 kt + 1} = \int \frac{dt}{{(t - k)}^{2} - (k^{2} - 1)} = \frac{- 1}{\sqrt{k^{2} - 1}} argtanh (\frac{t - k}{\sqrt{k^{2} - 1}}), ∣ k ∣> 1$ $I = \frac{2}{k - 1} - \frac{2}{k + 1} (\frac{k}{k - 1}) + \frac{4 \sqrt{1 - k^{2}}}{{(1 - k^{2})}^{2}} [\arctan (\sqrt{\frac{1 - k}{1 + k}}) + \arctan (\frac{k + 1}{\sqrt{1 - k^{2}}})], ∣ k ∣< 1$ $= \frac{2}{k - 1} - \frac{2}{k + 1} (\frac{k}{k - 1}) + \frac{2 \sqrt{k^{2} - 1}}{{(k^{2} - 1)}^{2}} \ln ∣ \frac{\sqrt{k^{2} - 1} - k + 1}{\sqrt{k^{2} - 1} + k - 1} ∣, ∣ k ∣> 1$
	Commented by Ar Brandon last updated on 13/Jun/21

	$Please check!$
	Answered by mathmax by abdo last updated on 13/Jun/21
	$f(t)=∫_0 ^(2π) (dx/(t−ksinx)) ⇒f^′ (t)=−∫_0 ^(2π) (dx/((t−ksinx)^2 )) ⇒ ∫_0 ^(2π) (dx/((1−ksinx)^2 ))=−f^′ (1) changement e^(ix) =t give f(t)=∫_(∣z∣=1) (dz/(iz(t−k((z−z^(−1) )/(2i))))) =∫_(∣z∣=1) ((2idz)/(iz(2it−kz+kz^(−1) ))) =∫ ((2dz)/(2itz−kz^2 +k)) =∫_(∣z∣=1) ((−2dz)/(kz^2 −2itz −k)) let Φ(z)=((−2)/(kz^2 −2itz −k)) poles of Φ? Δ^′ =−t^2 +k^2 case 1 −t^2 +k^2 <0 ⇒∣k∣<∣t∣ ⇒z_1 =((it+i(√(k^2 −t^2 )))/k) z_2 =((it−i(√(k^2 −t^2 )))/k) ∣z_1 ∣=(1/(∣k∣))(√(t^2 +k^2 −t^2 ))=1 also ∣z_2 ∣=1 and Φ(z)=((−2)/(k(z−z_1 )(z−z_2 ))) ∫_R Φ(z)dz =2iπ{Res(Φ,z_1 )+Res(Φ,z_2 )} Res(Φ,z_1 )=((−2)/(k(z_1 −z_2 ))) Res(Φ,z_2 ) =((−2)/(k(z_2 −z_1 ))) ⇒Σ Res=0=∫_R Φ(z)dz ⇒f(t)=0 ⇒f^′ (t)=0 case 2 −t^2 +k^2 >0 ⇒∣k∣>∣t∣ ⇒z_1 =((it+(√(k^2 −t^2 )))/k) z_2 =((it−(√(k^2 −t^2 )))/k) ∣z_1 ∣ =(1/(∣k∣))(√(t^2 +k^2 −t^2 ))=1 and ∣z_2 ∣=1 ∫_(∣z∣=1) Φ(z)dz =2iπ{Res(Φ,z_1 )+Res(Φ,z_2 )} but we get Res(Φ,z_1 )=−Res(Φ,z_2 ) ⇒Σ Res=0⇒∫_(∣z∣) Φ(z)dz=0 ⇒ f(t)=0 ⇒f^′ (t)=0$
	$f (t) = \int_{0}^{2 π} \frac{dx}{t - ksinx} \Rightarrow f^{^{'}} (t) = - \int_{0}^{2 π} \frac{dx}{{(t - ksinx)}^{2}} \Rightarrow$ $\int_{0}^{2 π} \frac{dx}{{(1 - ksinx)}^{2}} = - f^{^{'}} (1) changement e^{ix} = t give$ $f (t) = \int_{∣ z ∣= 1} \frac{dz}{iz (t - k \frac{z - z^{- 1}}{2 i})} = \int_{∣ z ∣= 1} \frac{2 idz}{iz (2 it - kz + {kz}^{- 1})}$ $= \int \frac{2 dz}{2 itz - {kz}^{2} + k} = \int_{∣ z ∣= 1} \frac{- 2 dz}{{kz}^{2} - 2 itz - k}$ $let Φ (z) = \frac{- 2}{{kz}^{2} - 2 itz - k} poles of Φ ?$ $Δ^{^{'}} = - t^{2} + k^{2}$ $case 1 - t^{2} + k^{2} < 0 \Rightarrow∣ k ∣<∣ t ∣ \Rightarrow z_{1} = \frac{it + i \sqrt{k^{2} - t^{2}}}{k}$ $z_{2} = \frac{it - i \sqrt{k^{2} - t^{2}}}{k}$ $∣ z_{1} ∣= \frac{1}{∣ k ∣} \sqrt{t^{2} + k^{2} - t^{2}} = 1 also ∣ z_{2} ∣= 1 and Φ (z) = \frac{- 2}{k (z - z_{1}) (z - z_{2})}$ $\int_{R} Φ (z) dz = 2 i π {Res (Φ, z_{1}) + Res (Φ, z_{2})}$ $Res (Φ, z_{1}) = \frac{- 2}{k (z_{1} - z_{2})}$ $Res (Φ, z_{2}) = \frac{- 2}{k (z_{2} - z_{1})} \Rightarrow Σ Res = 0 = \int_{R} Φ (z) dz \Rightarrow f (t) = 0 \Rightarrow f^{^{'}} (t) = 0$ $case 2 - t^{2} + k^{2} > 0 \Rightarrow∣ k ∣>∣ t ∣ \Rightarrow z_{1} = \frac{it + \sqrt{k^{2} - t^{2}}}{k}$ $z_{2} = \frac{it - \sqrt{k^{2} - t^{2}}}{k}$ $∣ z_{1} ∣ = \frac{1}{∣ k ∣} \sqrt{t^{2} + k^{2} - t^{2}} = 1 and ∣ z_{2} ∣= 1$ $\int_{∣ z ∣= 1} Φ (z) dz = 2 i π {Res (Φ, z_{1}) + Res (Φ, z_{2})}$ $but we get Res (Φ, z_{1}) = - Res (Φ, z_{2}) \Rightarrow Σ Res = 0 \Rightarrow \int_{∣ z ∣} Φ (z) dz = 0 \Rightarrow$ $f (t) = 0 \Rightarrow f^{^{'}} (t) = 0$
	Commented by Ar Brandon last updated on 13/Jun/21

	$Hum . . . Th \overset{´}{e o r} \overset{`}{e m e} de R \overset{´}{e s i d u s} .$ $Thanks Sir! Grateful!$
	Commented by mathmax by abdo last updated on 13/Jun/21

	$you are welcome$
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