0-pi-d-a-cos-2-a-gt-1-

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

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 \dots

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 \dots

Commented by arcana last updated on 26/Jul/19

de nuevo gracias

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