calculate-0-2pi-x-sin-d-x-2-2x-sin-1-2-

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

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 .