∫_0 ^π (dθ/((a+cosθ)^2 )), a>1


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Question Number 65100 by arcana last updated on 25/Jul/19

$\int_{0}^{π} \frac{d θ}{{(a + c o s θ)}^{2}}, a > 1$
Commented by ~ À ® @ 237 ~ last updated on 25/Jul/19

$l e t g (a) = \int_{0}^{π} \frac{d θ}{a + c o s θ} t h e n \int_{0}^{π} \frac{d θ}{{(a + c o s θ)}^{2}} = - \frac{d g}{d a} s o l e t f i n d g (a)$ $g (a) = \int_{0}^{π} \frac{d θ}{(a - 1) + 2 {c o s}^{2} (\frac{θ}{2})} = \int_{0}^{\frac{π}{2}} \frac{2 d x}{(a - 1) [{c o s}^{2} x + {s i n}^{2} x] + 2 {c o s}^{2} x} = 2 \int_{0}^{\frac{π}{2}} \frac{d x}{(a + 1) {c o s}^{2} x + (a - 1) {s i n}^{2} x}$ $t h e n$ $\frac{g (a)}{2} = \int_{0}^{\frac{π}{2}} \frac{\frac{1}{{c o s}^{2} x} d x}{(a + 1) + (a - 1) {t a n}^{2} x}$ $w e h a v e a > 1 \Rightarrow (a + 1) + (a - 1) {t a n}^{2} x = (a + 1) [1 + {(\sqrt{\frac{a - 1}{a + 1}} t a n x)}^{2}]$ $\frac{g (a)}{2} = \frac{1}{\sqrt{a^{2} - 1}} \int_{0}^{\frac{π}{2}} \frac{\sqrt{\frac{a - 1}{a + 1}} (1 + {t a n}^{2} x) d x}{1 + {(\sqrt{\frac{a - 1}{a + 1}} t a n x)}^{2}}$ $S o w e g e t$ $g (a) = \frac{2}{\sqrt{a^{2} - 1}} {[a r c t a n (\sqrt{\frac{a - 1}{a + 1}} t a n x)]}_{0}^{\frac{π}{2}} = \frac{π}{\sqrt{a^{2} - 1}}$ $t h e n \frac{d g}{d a} = π . (\frac{\frac{- 2 a}{2 \sqrt{a^{2} - 1}}}{a^{2} - 1}) = \frac{- π a}{{(a^{2} - 1)}^{\frac{3}{2}}}$ $t h a t l e a d s u s t o$ $\int_{0}^{π} \frac{d θ}{{(a + c o s θ)}^{2}} = \frac{π a}{{(a^{2} - 1)}^{\frac{3}{2}}}$
Commented by turbo msup by abdo last updated on 25/Jul/19
$let ϕ(a) =∫_0 ^π (dθ/(a+cosθ)) we have ϕ^′ (a)=−∫_0 ^π (dθ/((a+cosθ)^2 )) ⇒ ∫_0 ^π (dθ/((a+cosθ)^2 )) =−ϕ^′ (a) ϕ(a) =_(tan((θ/2))=x) ∫_0 ^∞ (1/(a+((1−x^2 )/(1+x^2 )))) ((2dx)/(1+x^2 )) =∫_0 ^∞ ((2dx)/(a+ax^2 +1−x^2 )) =∫_0 ^∞ ((2dx)/((a−1)x^2 +a+1)) =(2/(a−1)) ∫_0 ^∞ (dx/(x^2 +((a+1)/(a−1)))) =_(x =(√((a+1)/(a−1)))u) (2/(a−1)) ∫_0 ^∞ (1/(((a+1)/(a−1))(1+u^2 )))(√((a+1)/(a−1)))du =(2/(a+1)) ((√(a+1))/(√(a−1))) ∫_0 ^∞ (du/(1+u^2 )) =(2/(√(a^2 −1))) (π/2) =(π/(√(a^2 −1))) ⇒ ϕ^′ (a) =π{(a^2 −1)^(−(1/2)) }^′ =π(−(1/2))(2a)(a^2 −1)^(−(3/2)) =((−πa)/((a^2 −1)(√(a^2 −1)))) ⇒ ∫_0 ^π (dθ/((a+cosθ)^2 )) =((πa)/((a^2 −1)(√(a^2 −1)))) if a>1$
$l e t φ (a) = {\int^{π}}_{0} \frac{d θ}{a + c o s θ} w e h a v e$ $φ^{^{'}} (a) = - \int_{0}^{π} \frac{d θ}{{(a + c o s θ)}^{2}} \Rightarrow$ $\int_{0}^{π} \frac{d θ}{{(a + c o s θ)}^{2}} = - φ^{^{'}} (a)$ $φ (a) =_{t a n (\frac{θ}{2}) = x} \int_{0}^{\infty} \frac{1}{a + \frac{1 - x^{2}}{1 + x^{2}}} \frac{2 d x}{1 + x^{2}}$ $= \int_{0}^{\infty} \frac{2 d x}{a + {a x}^{2} + 1 - x^{2}} = \int_{0}^{\infty} \frac{2 d x}{(a - 1) x^{2} + a + 1}$ $= \frac{2}{a - 1} \int_{0}^{\infty} \frac{d x}{x^{2} + \frac{a + 1}{a - 1}}$ $=_{x = \sqrt{\frac{a + 1}{a - 1}} u} \frac{2}{a - 1} \int_{0}^{\infty} \frac{1}{\frac{a + 1}{a - 1} (1 + u^{2})} \sqrt{\frac{a + 1}{a - 1}} d u$ $= \frac{2}{a + 1} \frac{\sqrt{a + 1}}{\sqrt{a - 1}} \int_{0}^{\infty} \frac{d u}{1 + u^{2}}$ $= \frac{2}{\sqrt{a^{2} - 1}} \frac{π}{2} = \frac{π}{\sqrt{a^{2} - 1}} \Rightarrow$ $φ^{^{'}} (a) = π {{(a^{2} - 1)}^{- \frac{1}{2}}}^{^{'}}$ $= π (- \frac{1}{2}) (2 a) {(a^{2} - 1)}^{- \frac{3}{2}}$ $= \frac{- π a}{(a^{2} - 1) \sqrt{a^{2} - 1}} \Rightarrow$ $\int_{0}^{π} \frac{d θ}{{(a + c o s θ)}^{2}} = \frac{π a}{(a^{2} - 1) \sqrt{a^{2} - 1}}$ $i f a > 1$
Commented by arcana last updated on 25/Jul/19

$Gracias$
Commented by arcana last updated on 25/Jul/19

$Gracias$
Commented by turbo msup by abdo last updated on 25/Jul/19

$w h e r e a r e y o u f r o m a r c a n a . . .$
Commented by arcana last updated on 25/Jul/19

$Colombia but i can understand coments .$ $Muchas gracias$
Commented by mathmax by abdo last updated on 25/Jul/19

$y o u a r e m o s t w e l c o m e i n t h a t p l a t f o r m . . .$
Commented by arcana last updated on 26/Jul/19

$de nuevo gracias$
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