calculate ∫_0 ^(2π) (((x−sinθ)dθ)/((x^2 −2x sinθ +1)^2 ))


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Question Number 74514 by mathmax by abdo last updated on 25/Nov/19

$c a l c u l a t e \int_{0}^{2 π} \frac{(x - s i n θ) d θ}{{(x^{2} - 2 x s i n θ + 1)}^{2}}$
Commented by mathmax by abdo last updated on 26/Nov/19

$w e h a v e p r o v e d t h a t \int_{0}^{π} \frac{d t}{x^{2} - 2 x s i n (2 t) + 1} = \frac{π}{1 - x^{2}} i f ∣ x ∣< 1 a n d$ $= \frac{π}{x^{2} - 1} i f ∣ x ∣> 1 l e t f (x) = \int_{0}^{π} \frac{d t}{x^{2} - 2 x s i n (2 t) + 1} \Rightarrow$ $f (x) =_{2 t = θ} \int_{0}^{2 π} \frac{d θ}{2 (x^{2} - 2 x s i n θ + 1)} \Rightarrow 2 f^{^{'}} (x) = - \int_{0}^{2 π} \frac{2 x - 2 s i n θ}{(x^{2} {(2 x s i n θ + 1)}^{2}} d θ \Rightarrow$ $\int_{0}^{2 π} \frac{(x - s i n θ)}{{(x^{2} - 2 x s i n θ + 1)}^{2}} d θ = - f^{^{'}} (x)$ $∣ x ∣< 1 \Rightarrow f^{^{'}} (x) = - π \frac{- 2 x}{{(1 - x^{2})}^{2}} = \frac{2 π x}{{(1 - x^{2})}^{2}}$ $∣ x ∣> 1 \Rightarrow f^{^{'}} (x) = - \frac{π (2 x)}{{(x^{2} - 1)}^{2}} = \frac{- 2 π x}{{(x^{2} - 1)}^{2}} s o t h e v a l u e o f t h i s i n t e g r a l$ $i s k n o w n .$
	Answered by mind is power last updated on 25/Nov/19
	$use provious Quation f(x)=∫_0 ^π (dθ/(x^2 −2sin(2θ)x+1)) u=2θ f(x)=∫_0 ^(2π) (du/(2(x^2 −2sin(u)x+1))) ⇒2f(x)=∫_0 ^(2π) (du/(x^2 −2sin(u)x+1))= { ((((πx)/(x^2 −1)), x>1)),(((π/(1−x^2 )), 0<x<1)) :} 2f′(x)=∫_0 ^(2π) ((−2x+2sin(u))/((x^2 −2xsin(u)+1)^2 )) ⇒−f′(x)=∫_0 ^(2π) ((x−sin(u))/((x^2 −2xsin(u)+1)^2 ))du f′(x)=π.(((−x^2 −1)/((x^2 +1)^2 ))) 0≤x<1 f′(x)=((2xπ)/((1−x^2 )^2 )),x>1$
	$u s e p r o v i o u s Q u a t i o n$ $f (x) = \int_{0}^{π} \frac{d θ}{x^{2} - 2 s i n (2 θ) x + 1}$ $u = 2 θ$ $f (x) = \int_{0}^{2 π} \frac{d u}{2 (x^{2} - 2 s i n (u) x + 1)}$ $\Rightarrow 2 f (x) = \int_{0}^{2 π} \frac{d u}{x^{2} - 2 s i n (u) x + 1} = {\begin{cases} \frac{π x}{x^{2} - 1}, x > 1 \\ \frac{π}{1 - x^{2}}, 0 < x < 1 \end{cases}$ $2 f^{'} (x) = \int_{0}^{2 π} \frac{- 2 x + 2 s i n (u)}{{(x^{2} - 2 x s i n (u) + 1)}^{2}}$ $\Rightarrow - f^{'} (x) = \int_{0}^{2 π} \frac{x - s i n (u)}{{(x^{2} - 2 x s i n (u) + 1)}^{2}} d u$ $f^{'} (x) = π . (\frac{- x^{2} - 1}{{(x^{2} + 1)}^{2}}) 0 ⩽ x < 1$ $f^{'} (x) = \frac{2 x π}{{(1 - x^{2})}^{2}}, x > 1$
	Commented by mathmax by abdo last updated on 26/Nov/19

	$t h a n k y o u s i r .$
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