calculate ∫_0 ^(4π) (dθ/((x^2 +2x cosθ +1)^2 ))


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Question Number 141222 by mathmax by abdo last updated on 16/May/21

$calculate \int_{0}^{4 π} \frac{d θ}{{(x^{2} + 2 x \cos θ + 1)}^{2}}$
	Answered by Ar Brandon last updated on 16/May/21
	$Ω=∫_0 ^(4π) (dθ/((x^2 +2xcosθ+1)^2 ))=4∫_0 ^π (dθ/((x^2 +2xcosθ+1)^2 )), t=tan(θ/2) =4∫_0 ^∞ ((2dt)/((x^2 +2x∙((1−t^2 )/(1+t^2 ))+1)^2 ))∙(1/(1+t^2 )) =8∫_0 ^∞ ((t^2 +1)/((x^2 (1+t^2 )+2x(1−t^2 )+1+t^2 )^2 ))dt =8∫_0 ^∞ ((t^2 +1)/(((x^2 −2x+1)t^2 +(x^2 +2x+1))^2 ))dt =8∫_0 ^∞ ((t^2 +1)/(((x−1)^2 t^2 +(x+1)^2 )^2 ))=(8/((x−1)^4 ))∫_0 ^∞ ((t^2 +1)/((t^2 +(((x+1)/(x−1)))^2 )^2 )) =(8/((x−1)^4 )){∫_0 ^∞ ((t^2 +(((x+1)/(x−1)))^2 )/((t^2 +(((x+1)/(x−1)))^2 )^2 ))dt+∫_0 ^∞ ((1−(((x+1)/(x−1)))^2 )/((t^2 +(((x+1)/(x−1)))^2 )^2 ))dt} f(a)=∫_0 ^∞ (dt/(t^2 +a^2 ))=(π/(2a))⇒f ′(a)=−∫_0 ^∞ ((2a)/((t^2 +a^2 )^2 ))dt=−(π/(2a^2 )) ⇒∫_0 ^∞ (dt/((t^2 +a^2 )^2 ))=(π/(4a^3 )) Ω=(8/((x−1)^4 )){(π/(2(((x+1)/(x−1)))))+(1−(((x+1)/(x−1)))^2 )×(π/(4(((x+1)/(x−1)))^3 ))}$
	$Ω = \int_{0}^{4 π} \frac{d θ}{{(x^{2} + 2 xcos θ + 1)}^{2}} = 4 \int_{0}^{π} \frac{d θ}{{(x^{2} + 2 xcos θ + 1)}^{2}}, t = \tan \frac{θ}{2}$ $= 4 \int_{0}^{\infty} \frac{2 dt}{{(x^{2} + 2 x \cdot \frac{1 - t^{2}}{1 + t^{2}} + 1)}^{2}} \cdot \frac{1}{1 + t^{2}}$ $= 8 \int_{0}^{\infty} \frac{t^{2} + 1}{{(x^{2} (1 + t^{2}) + 2 x (1 - t^{2}) + 1 + t^{2})}^{2}} dt$ $= 8 \int_{0}^{\infty} \frac{t^{2} + 1}{{((x^{2} - 2 x + 1) t^{2} + (x^{2} + 2 x + 1))}^{2}} dt$ $= 8 \int_{0}^{\infty} \frac{t^{2} + 1}{{({(x - 1)}^{2} t^{2} + {(x + 1)}^{2})}^{2}} = \frac{8}{{(x - 1)}^{4}} \int_{0}^{\infty} \frac{t^{2} + 1}{{(t^{2} + {(\frac{x + 1}{x - 1})}^{2})}^{2}}$ $= \frac{8}{{(x - 1)}^{4}} {\int_{0}^{\infty} \frac{t^{2} + {(\frac{x + 1}{x - 1})}^{2}}{{(t^{2} + {(\frac{x + 1}{x - 1})}^{2})}^{2}} dt + \int_{0}^{\infty} \frac{1 - {(\frac{x + 1}{x - 1})}^{2}}{{(t^{2} + {(\frac{x + 1}{x - 1})}^{2})}^{2}} dt}$ $f (a) = \int_{0}^{\infty} \frac{dt}{t^{2} + a^{2}} = \frac{π}{2 a} \Rightarrow f^{'} (a) = - \int_{0}^{\infty} \frac{2 a}{{(t^{2} + a^{2})}^{2}} dt = - \frac{π}{{2 a}^{2}}$ $\Rightarrow \int_{0}^{\infty} \frac{dt}{{(t^{2} + a^{2})}^{2}} = \frac{π}{{4 a}^{3}}$ $Ω = \frac{8}{{(x - 1)}^{4}} {\frac{π}{2 (\frac{x + 1}{x - 1})} + (1 - {(\frac{x + 1}{x - 1})}^{2}) \times \frac{π}{4 {(\frac{x + 1}{x - 1})}^{3}}}$
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