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Question Number 41913 by math khazana by abdo last updated on 15/Aug/18

l e t f (a) = \int_{0}^{π} \frac{x}{1 + a c o s x} d x

1) f i n d f (a)

2) c a l c u l a t e \int_{0}^{π} \frac{x}{1 + 2 c o s x} d x a n d \int_{0}^{π} \frac{x}{1 - 2 c o s x} d x

3) c a l c u l a t e \int_{0}^{π} \frac{x c o s x}{{(1 + a c o s x)}^{2}} d x

4) f i n d t h e v a l u e o f \int_{0}^{π} \frac{x c o s x}{{(1 + 2 c o s x)}^{2}} d x .

Answered by maxmathsup by imad last updated on 16/Aug/18

1) w e h a v e f (a) = \int_{0}^{π} \frac{x}{1 + a \frac{2 t a n (\frac{x}{2})}{1 + {t a n}^{2} (\frac{x}{2})}} d x c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e

f (a) = \int_{0}^{\infty} \frac{2 a r c t a n (t)}{1 + \frac{2 a t}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = \int_{0}^{\infty} \frac{4 a r c t a n (t)}{1 + t^{2} + 2 a t} d t = \int_{0}^{\infty} \frac{4 a r c t a n (t)}{t^{2} + 2 a t + 1} d t

= 4 \int_{0}^{\infty} \frac{a r c t a n (t)}{t^{2} + 2 a t + 1} d t = w (1) w i t h w (α) = \int_{0}^{\infty} \frac{a r c t a n (α t)}{t^{2} + 2 a t + 1} d t w e h a v e

w^{^{'}} (α) = \int_{0}^{\infty} \frac{t}{(1 + α^{2} t^{2}) (t^{2} + 2 a t + 1)} d t =_{α t = u} \int_{0}^{\infty} \frac{u}{α (1 + u^{2}) (\frac{u^{2}}{α^{2}} + 2 a \frac{u}{α} + 1)} \frac{d u}{α}

= \int_{0}^{\infty} \frac{u d u}{α^{2} (1 + u^{2}) (\frac{u^{2} + 2 α a u + α^{2}}{α^{2}})} = \int_{0}^{\infty} \frac{u d u}{(u^{2} + 1) (u^{2} + 2 α a u + α^{2})}

l e t d e c o m p o s e F (u) = \frac{u}{(u^{2} + 1) (u^{2} + 2 α a u + α^{2})} (l e t t a k e α > 0)

r o o t s o f u^{2} + 2 α a u + α^{2}

Δ^{^{'}} = α^{2} a^{2} - α^{2} = α^{2} (a^{2} - 1)

c a s e 1) ∣ a ∣> 1 \Rightarrow u_{1} = - α a + ∣ α ∣ \sqrt{a^{2} - 1} a n d u_{2} = - α a - ∣ α ∣ \sqrt{a^{2} - 1}

F (u) = \frac{a}{u - u_{1}} + \frac{b}{u - u_{2}} + \frac{c u + d}{u^{2} + 1} = \frac{u}{(u - u_{1}) (u - u_{2}) (u^{2} + 1)}

a = \frac{u_{1}}{(u_{1} - u_{2}) (u_{1}^{2} + 1)} = \frac{∣ α ∣ \sqrt{a^{2} - 1} - α a}{2 ∣ α ∣ \sqrt{a^{2} - 1} ({(∣ α ∣ \sqrt{a^{2} - 1} - α a)}^{2} + 1)} = \frac{α (\sqrt{a^{2} - 1} - a)}{2 α \sqrt{a^{2} - 1} (α^{2} {(\sqrt{a^{2} - 1} - a)}^{2} + 1)}

= \frac{\sqrt{a^{2} - 1} - a}{2 \sqrt{a^{2} - 1} {α^{2} {(\sqrt{a^{2} - 1} - a)}^{2} + 1}}

b = \frac{u_{2}}{(u_{2} - u_{1}) (u_{2}^{2} + 1)} = \frac{- α (a + \sqrt{a^{2} - 1})}{- 2 α \sqrt{a^{2} - 1} {α^{2} {(a + \sqrt{a^{2} - 1})}^{2} + 1}}

{l i m}_{u \to + \infty} u F (u) = 0 = a + b + c \Rightarrow c = - a - b \Rightarrow

F (u) = \frac{a}{u - u_{1}} + \frac{b}{u - u_{2}} + \frac{(- a - b) u + d}{u^{2} + 1}

F (0) = 0 = - \frac{a}{u_{1}} - \frac{b}{u_{2}} + d \Rightarrow d = \frac{a}{u_{1}} + \frac{b}{u_{2}} \Rightarrow

F (u) = \frac{a}{u - u_{1}} + \frac{b}{u - u_{2}} + \frac{(- a - b) u + \frac{a}{u_{1}} + \frac{b}{u_{2}}}{u^{2} + 1} \Rightarrow

\int_{0}^{\infty} F (u) d u = {[a l n ∣ u - u_{1} ∣ + b l n ∣ u - u_{2} ∣]}_{0}^{+ \infty} - {[\frac{a + b}{2} l n (u^{2} + 1)]}_{0}^{+ \infty}

+ (\frac{a}{u_{1}} + \frac{b}{u_{2}}) {[a r c t a n (u)]}_{0}^{+ \infty}

= {[l n (∣ u - u_{1} ∣^{a} ∣ u - u_{2} ∣^{b}) - l n ({(u^{2} + 1)}^{\frac{a + b}{2}})]}_{0}^{+ \infty} + \frac{π}{2} {\frac{a}{u_{1}} + \frac{b}{u_{2}}} \dots b e c o n t i n u e d \dots